# general chemistry

#### olygt

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If a molecule is composed of only two elements (X and Y) and if X and Y combine in equal mass quantities, and if Y is less than twice as heavy as X, which of the following molecular formulas is NOT possible?

A) XY
B) XY2
C) X3Y2
D) X3Y

The answer is D, but I would think the answer is B since the molecular mass of X would be equal to or greater than Y, and B can't be possible because Y is bigger.

#### Chemist0157

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I thought the same thing at first, but then I reread the problem.

Y isn't necessarily bigger than X just less than twice the size of X.

X3Y isn't possible because that means that Y is 3 times bigger than X.

XY2 is possible because that means that Y is half the size of X.

Pretty tricky, but it makes sense according to the parameters.

#### olygt

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I thought the same thing at first, but then I reread the problem.

Y isn't necessarily bigger than X just less than twice the size of X.

X3Y isn't possible because that means that Y is 3 times bigger than X.

XY2 is possible because that means that Y is half the size of X.

Pretty tricky, but it makes sense according to the parameters.

ah ha...I think I got it, thanks.

#### olygt

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Can somebody clarify this statement, it's a bit confusing: The molality of a solution does not change with temperature, so it is often used to determine a change in the solution's temperature when the change depends on concentration. Notable examples of this include boiling point and freezing pt. depression.

#### tncekm

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Molality is moles/mass solvent. Mass of solvent never changes with temperature, unlike volume.

I.e. if you heat H2O from 99 C to 101 C its volume has changed drastically as it has gone from the liquid to gas phase. However, its mass is still the same.

So, if you had (theoretical) 2L solution of CO2 in liquid water at [0.000001]M confined to a 10L container if you heated the solution from 99 C to 101C it would do the following:

In molarity (M)
Start (99, liquid): [0.000001]mol CO2/L water x 2L water = 0.000002 mol CO2
Finish (101, gas): [0.000002]/10L = [0.0000001]M (Remember, a gas will fill its entire container, so the new volume is 10L).

In molality (m)
2L water = 2kg
Start (99, liquid): [0.000001]mol CO2/Kg = [0.000001] molal
Finish (101, gas): We still have only 2kg water, even though it occupies more volume. So, its still [0.000001] molal

In conclusion, the molar concentration (M) changed from 10^-6 to 10^-7 Molar during this heating process and phase change, and the molal solution (m) stayed the same at 10^-6 molal.

#### stupidkid30

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chimyyanka wakawaka chemistry sucks but yeah I agree with the above poster, thats exactly what I was going to say, I mean really, word per word.

#### olygt

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How does the second ionization energy of sodium compare to the second ionization energ of magnesium?

A) Na has a greater second ionization energy b/c it has the greater effective nuclear charge
B) Mg has a greater second ionization energy b/c it has the greater effective nuclear charge
C) Na has a greater second ionization energy b/c it has the smaller electronic shell
D) Mg has a greater second ionization energy b/c it has the smaller electronic shell

I automatically crossed out C and D being that they both end up in the 2p shell b/c Na ends up with 9 electrons and 10 electrons. This left me chosing B as my answer which is wrong, the answer is C. How...I haven't a clue. Here is the books explanation: "For sodium, an octet is obtained by losing the first electron, thus the second electron lost drastically destabilizes the electron cloud. This makes teh second ionization energy very high. For Mg, an octet is obtained by losing the first and second electron, thus the second electron lost stabilizes the electron cloud. (ALL THAT I UNDERSTAND). This makes the second ionization energy very low for Mg, eliminated B and D. B/w we are talking about shell stability, the best answer is choice C.

#### RSAgator

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Well I'm guessing Mg/Na ions. For Na, the first electron is removed relatively easily because doing so results in a full octet. The second electron is incredibly difficult to remove however. It's the equivalent of trying to remove an electron from one of the halogens, it already has a full octet so it's very stable.

#### RPedigo

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Well I'm guessing Mg/Na ions. For Na, the first electron is removed relatively easily because doing so results in a full octet. The second electron is incredibly difficult to remove however. It's the equivalent of trying to remove an electron from one of the halogens, it already has a full octet so it's very stable.
Excellent explanation, but just to clarify, the second electron removal from sodium is the equivalent of trying to remove an electron from one of the noble gases, not one of the halogens. Na+ is isoelectronic with Ne. Otherwise, great job #### eeyoreDO

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If a molecule is composed of only two elements (X and Y) and if X and Y combine in equal mass quantities, and if Y is less than twice as heavy as X, which of the following molecular formulas is NOT possible?

A) XY
B) XY2
C) X3Y2
D) X3Y

The answer is D, but I would think the answer is B since the molecular mass of X would be equal to or greater than Y, and B can't be possible because Y is bigger.
I finally figured out how to do this question!!

You can assign numerical values to the X and Y values and see if they match up.

The answer, D, is X3Y. That means that X3 = Y. For example, X could be 1 and Y could be 3.

"Y is less than twice as heavy as X" means Y < 2X. In the numbers I assigned (x = 1 and y = 3), this wouldn't hold true.

Y < 2X
(3) < 2(1)

Woohoo!

#### Kaustikos

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How does the second ionization energy of sodium compare to the second ionization energ of magnesium?

A) Na has a greater second ionization energy b/c it has the greater effective nuclear charge
B) Mg has a greater second ionization energy b/c it has the greater effective nuclear charge
C) Na has a greater second ionization energy b/c it has the smaller electronic shell
D) Mg has a greater second ionization energy b/c it has the smaller electronic shell

I automatically crossed out C and D being that they both end up in the 2p shell b/c Na ends up with 9 electrons and 10 electrons. This left me chosing B as my answer which is wrong, the answer is C. How...I haven't a clue. Here is the books explanation: "For sodium, an octet is obtained by losing the first electron, thus the second electron lost drastically destabilizes the electron cloud. This makes teh second ionization energy very high. For Mg, an octet is obtained by losing the first and second electron, thus the second electron lost stabilizes the electron cloud. (ALL THAT I UNDERSTAND). This makes the second ionization energy very low for Mg, eliminated B and D. B/w we are talking about shell stability, the best answer is choice C.
Coudl someone explain why it is not A? My reasoning is that effective charge deals with protons vs valence electrons and that with the removal of the 2 electrons, the effective charge is affected. But is it because they say the effective charge increase that it is wrong? #### eeyoreDO

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damn, I don't know about that effective nuclear charge

#### BerkReviewTeach

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Excellent explanation, but just to clarify, the second electron removal from oxygen is the equivalent of trying to remove an electron from one of the noble gases, not one of the halogens. Na+ is isoelectronic with Ne. Otherwise, great job I think you mean an alkaline earth metal rather than a chalcogen such as oxygen.

#### BerkReviewTeach

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Coudl someone explain why it is not A? My reasoning is that effective charge deals with protons vs valence electrons and that with the removal of the 2 electrons, the effective charge is affected. But is it because they say the effective charge increase that it is wrong? In dealing with the second ionization of sodium, the electron is being removed from the second shell (1s22s22p6 --> 1s22s22p5). In dealing with the second ionization of magnesium, the electron is being removed from the third shell (1s22s22p63s1 --> 1s22s22p6). This means that the electron is coming from the second shell for sodium and the third shell for magnesium. While the effective nuclear charges are different (greater for Na than Mg), the better answer is shells, because that (a filled shell) accounts for noble gas configuration.

#### RPedigo

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I think you mean an alkaline earth metal rather than a chalcogen such as oxygen.
Oops. I was too tired last night I guess haha. But I meant second removal from an alkali metal, not alkaline earth metal. Na+ is isoelectronic with Ne.

#### olygt

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When do know if you should ignore x or consider x when working out equilibrium shifts. I'm reading that if Keq is less than 1.0 hardly anything shift over to the product side, so x can be ignored. How do you know that hardly anything shifts over? Here is an example of when x is considered:

At 92.2 C, the Kp for the following reaction is 0.2000 atm-1. If you were to place exactly 0.200 atm of N2O4 into a 1.00 L vessel, what would the partial pressure of NO2 be once equilibrium was established?
2NO2 --------> N2O4

A) 0.025 atm NO2
B) 0.200 atm NO2
C) 0.350 atm NO2
D) 0.400 atm NO2

Can anybody explain to me the logic in the explanation, without even working any math out for this problem? I probably can do this math, but that would take about five minutes to work out and of course the MCAT doesn't give us that time.

Here's the explanation: In this case Keq is less than 1 and the rxn starts with all products. The value of x is going to be sig. (more than half shifts over). If half of the 0.200 atm of N2O4 sifts over, than the partial pressure of nitrogen dioxide is 0.200 atm. Considering that more than half of the N2O4 is going to shift, the value of NO2 is greater than 0.200 atm. However, not all of the
N2O4 can shift over (which would result in 0.400 atm of NO2), so the answer must be less than 0.400 atm. Only choice C falls within the range of 0.200 and 0.400.

#### Kaustikos

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When do know if you should ignore x or consider x when working out equilibrium shifts. I'm reading that if Keq is less than 1.0 hardly anything shift over to the product side, so x can be ignored. How do you know that hardly anything shifts over? Here is an example of when x is considered:

At 92.2 C, the Kp for the following reaction is 0.2000 atm-1. If you were to place exactly 0.200 atm of N2O4 into a 1.00 L vessel, what would the partial pressure of NO2 be once equilibrium was established?
2NO2 --------> N2O4

A) 0.025 atm NO2
B) 0.200 atm NO2
C) 0.350 atm NO2
D) 0.400 atm NO2

Can anybody explain to me the logic in the explanation, without even working any math out for this problem? I probably can do this math, but that would take about five minutes to work out and of course the MCAT doesn't give us that time.

Here's the explanation: In this case Keq is less than 1 and the rxn starts with all products. The value of x is going to be sig. (more than half shifts over). If half of the 0.200 atm of N2O4 sifts over, than the partial pressure of nitrogen dioxide is 0.200 atm. Considering that more than half of the N2O4 is going to shift, the value of NO2 is greater than 0.200 atm. However, not all of the
N2O4 can shift over (which would result in 0.400 atm of NO2), so the answer must be less than 0.400 atm. Only choice C falls within the range of 0.200 and 0.400.
I'm at a lost as to what you mean by X and the significance of it. I look at Keq and compare the the value to rate of forward vs reverse. I must suck and need to reread this section, but the way I figured this was looking at the Keq and seeing the 0.200 and knowing that the rate of reverse is more favored/occurs more quickly but not in a completely dominant fashion.

I'm not giving an answer, but trying to solve this myself, btw.

So 0.200 rate means product formation occurs less than reactant formation and you have a higher overall concentration of reactants at equilibrium. But it doesn't occur completely (and i'm not 100% on this so late at night but it just doesn't seem to be that you'll have a complete conversion of either product/reactant because then it would go to completion) The 0.200 tell me that it's higher than half otherwise it would be closer to a 0.5 rate to get the 0.200 mole formation of NO2 (Remember 1 mole of N2O4 to 2 moles NO2) It's not 0.400 mole because that would be a complete conversion.

Think of it this way, if I am correct which I may not be, but a value less than 1 means the reverse reaction happens more quickly. The closer you get to 0, the more likely the product is going to occur over the reactant. A value under 0.500 means more than half of products become reactant. A value of over 0.500 and less than 1 means less than half of products.

Flip it and reverse meanie: jk, just reverse it) for values above 1 to get an understanding of how Keq affects product/reactant formation.

But what did you mean by the x shift? #### BerkReviewTeach

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BR General Chemistry Sample Question 3.9

When do know if you should ignore x or consider x when working out equilibrium shifts. I'm reading that if Keq is less than 1.0 hardly anything shift over to the product side, so x can be ignored. How do you know that hardly anything shifts over? Here is an example of when x is considered:

At 92.2 C, the Kp for the following reaction is 0.2000 atm-1. If you were to place exactly 0.200 atm of N2O4 into a 1.00 L vessel, what would the partial pressure of NO2 be once equilibrium was established?
2NO2 --------> N2O4

A) 0.025 atm NO2
B) 0.200 atm NO2
C) 0.350 atm NO2
D) 0.400 atm NO2

Can anybody explain to me the logic in the explanation, without even working any math out for this problem? I probably can do this math, but that would take about five minutes to work out and of course the MCAT doesn't give us that time.

Here's the explanation: In this case Keq is less than 1 and the rxn starts with all products. The value of x is going to be sig. (more than half shifts over). If half of the 0.200 atm of N2O4 sifts over, than the partial pressure of nitrogen dioxide is 0.200 atm. Considering that more than half of the N2O4 is going to shift, the value of NO2 is greater than 0.200 atm. However, not all of the
N2O4 can shift over (which would result in 0.400 atm of NO2), so the answer must be less than 0.400 atm. Only choice C falls within the range of 0.200 and 0.400.

Perhaps a distribution chart can help here. Let's consider the four answer choices.

&#183; &#183; &#183; &#183; &#183; &#183; &#183; &#183; NO2 &#183; &#183; &#183; &#183; N2O4
start &#183; &#183; &#183; &#183; &#183; 0 &#183; &#183; &#183; &#183; 0.200
choice A &#183; 0.025 &#183; &#183; &#183; &#183; 0.1875
choice B &#183; 0.200 &#183; &#183; &#183; &#183; 0.1000
choice C &#183; 0.350 &#183; &#183; &#183; &#183; 0.0250
choice D &#183; 0.400 &#183; &#183; &#183; &#183; 0

Choice A can't be equilibrium, because there are more products than reactants. Given that K = 0.20, there must be more NO2 than N2O4.

Choice B can't be equilibrium, because plugging into the K expression yields 0.1/(0.2)exp2 = 0.1/0.04 = 10/4 > 0.200. Given that K = 0.20, that can't be right.

Choice D can't be equilibrium, because there must be at least some product.

Only choice C remains standing, thus it must be the best answer.

As the answer explanation states (BR page 174 of General Chemistry I), this question is designed to demonstrate that you don't really need to employ a great deal of math to solve numerical-based questions in a multiple-choice format.

BTW, if it makes you feel any better, the questions you sk are the usual suspects that most of our students ask about. Getting those questions correct is far less important to your preparation than figuring out a logical way to solve the questions quickly. You are doing quite well from what it seems. Three chapters down and seven to go.

#### Kaustikos

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So how wrong was I? #### olygt

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I'm at a lost as to what you mean by X and the significance of it. I look at Keq and compare the the value to rate of forward vs reverse. I must suck and need to reread this section, but the way I figured this was looking at the Keq and seeing the 0.200 and knowing that the rate of reverse is more favored/occurs more quickly but not in a completely dominant fashion.

I'm not giving an answer, but trying to solve this myself, btw.

So 0.200 rate means product formation occurs less than reactant formation and you have a higher overall concentration of reactants at equilibrium. But it doesn't occur completely (and i'm not 100% on this so late at night but it just doesn't seem to be that you'll have a complete conversion of either product/reactant because then it would go to completion) The 0.200 tell me that it's higher than half otherwise it would be closer to a 0.5 rate to get the 0.200 mole formation of NO2 (Remember 1 mole of N2O4 to 2 moles NO2) It's not 0.400 mole because that would be a complete conversion.

Think of it this way, if I am correct which I may not be, but a value less than 1 means the reverse reaction happens more quickly. The closer you get to 0, the more likely the product is going to occur over the reactant. A value under 0.500 means more than half of products become reactant. A value of over 0.500 and less than 1 means less than half of products.

Flip it and reverse meanie: jk, just reverse it) for values above 1 to get an understanding of how Keq affects product/reactant formation.

But what did you mean by the x shift? Thanks Kaustikos. I attached a .jpeg file drawing out a reaction from BR considering the x and ignoring it. Considering it requires during calculations with the quadratic equation, of course we aren't expected to work it all out on the MCAT, but if we did have to consider the x then we would have to do it intuitively.

#### olygt

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Which of the following mixtures results in a buffered solution?

A) 10 mL 0.25 M NaOH + 10 mL 0.25 M H3CCO2H
B) 20 mL 0.25 M NaOH + 10 mL 0.25 M H3CCO2H
C) 10 mL 0.25 M NaOH + 20 mL 0.25 M H3CCO2H
D) 10 mL 0.25 M NaOH + 10 mL 0.25 M H3CCO2H

The answer is C, but wouldn't B also produce a buffer since that one adds 1/2 an equivalent of the acid? Maybe it's b/c the stronger one (either base or acid) is the one to be added as half an equivalent??? I just realized that as I'm writing this...just to make sure, is that right?

#### tncekm

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A buffered solution will be half acid, and half conjugate base.

Option C removes half the protons from H3CCO2H leaving half acid and half conjugate base.

Option B removes all of the protons from H3CCO2H and then dissociates (as a strong electrolyte) into Na+ and OH- ions with the other 10mLx0.25M, and you'd have an unbuffered basic solution. Half the equivalent of a weak acid mixed with one equivalent of strong base just means that all the protons are stripped from the acetic acid, and it can't act like a buffer because a buffer can both release and take on protons.

This highlights an important thing, too. The reason you want a buffer solution with a pKa = pH of your desired solution is because when the conjugate acids and bases are equal to each other the henderson hasselbach equation pH = pKa-log(A/HA) --> pH = pKa. This is the half equivalence point, the point where buffering is most efficient.

You're on the right track with your thinking though #### BerkReviewTeach

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A buffered solution will be half acid, and half conjugate base.

Option C removes half the protons from H3CCO2H leaving half acid and half conjugate base.

Option B removes all of the protons from H3CCO2H and then dissociates (as a strong electrolyte) into Na+ and OH- ions with the other 10mLx0.25M, and you'd have an unbuffered basic solution. Half the equivalent of a weak acid mixed with one equivalent of strong base just means that all the protons are stripped from the acetic acid, and it can't act like a buffer because a buffer can both release and take on protons.

This highlights an important thing, too. The reason you want a buffer solution with a pKa = pH of your desired solution is because when the conjugate acids and bases are equal to each other the henderson hasselbach equation pH = pKa-log(A/HA) --> pH = pKa. This is the half equivalence point, the point where buffering is most efficient.

You're on the right track with your thinking though YEAH!!! tncekm, you own buffers! This is one of those watermark questions in our book. Students who get this as smoothly as you just got it are set for buffers.

#### tncekm

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Thanks... now if only I could own the rest of the MCAT #### olygt

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I'm not sure how this answer is correct, but BR got (C) as the correct answer.

If a drug decomposes according to first-order kinetics, what is its concentration after one hour, if the halflife is 21 minutes.

A) 26.1% of its original concentration
B) 42.0 % of its original concentration
C) 13.3 % of its original concentration
D) 11.7% of its original concentration

I got D as my answer b/c after three half-lives at 12.5% the time is 63 minutes, not 60 minutes, so just after one hour shouldn't the half-life be less than the 12.5%? If the question wanted any percent after one hour than it could have been A, B or C

#### RPedigo

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I'm not sure how this answer is correct, but BR got (C) as the correct answer.

If a drug decomposes according to first-order kinetics, what is its concentration after one hour, if the halflife is 21 minutes.

A) 26.1% of its original concentration
B) 42.0 % of its original concentration
C) 13.3 % of its original concentration
D) 11.7% of its original concentration

I got D as my answer b/c after three half-lives at 12.5% the time is 63 minutes, not 60 minutes, so just after one hour shouldn't the half-life be less than the 12.5%? If the question wanted any percent after one hour than it could have been A, B or C
Right, so 12.5% is left after 63 minutes. It asked for an hour, or 60 minutes. You went too long, which means at 60 minutes, there must be MORE than 12.5% left. It took you 63 minutes to get to 12.5% so at 60 minutes there must have been >12.5% left. 13.3% is a reasonable answer. To get to 11.7%, answer choice D, it would take longer than an hour, and longer than 63 minutes too.

Think of it this way:
t = 00 minutes, 100% left
t = 21 minutes, 50% left
t = 42 minutes, 25% left
(t = 60 minutes is here, so between 25% and 12.5% left, but closer to 12.5%)
t = 63 minutes, 12.5% left

#### olygt

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Right, so 12.5% is left after 63 minutes. It asked for an hour, or 60 minutes. You went too long, which means at 60 minutes, there must be MORE than 12.5% left. It took you 63 minutes to get to 12.5% so at 60 minutes there must have been >12.5% left. 13.3% is a reasonable answer. To get to 11.7%, answer choice D, it would take longer than an hour, and longer than 63 minutes too.

Think of it this way:
t = 00 minutes, 100% left
t = 21 minutes, 50% left
t = 42 minutes, 25% left
(t = 60 minutes is here, so between 25% and 12.5% left, but closer to 12.5%)
t = 63 minutes, 12.5% left
ahh...ok, I guess I was looking at it from the wrong perspective. Got it now, though, thanks!

#### Livingapparatus

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If a molecule is composed of only two elements (X and Y) and if X and Y combine in equal mass quantities, and if Y is less than twice as heavy as X, which of the following molecular formulas is NOT possible?

A) XY
B) XY2
C) X3Y2
D) X3Y

The answer is D, but I would think the answer is B since the molecular mass of X would be equal to or greater than Y, and B can't be possible because Y is bigger.
this question is more of a word problem then a chemistry problem.

#### BerkReviewTeach

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this question is more of a word problem then a chemistry problem.
Welcome to the MCAT!

#### TJames

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If a molecule is composed of only two elements (X and Y) and if X and Y combine in equal mass quantities, and if Y is less than twice as heavy as X, which of the following molecular formulas is NOT possible?

A) XY
B) XY2
C) X3Y2
D) X3Y

The answer is D, but I would think the answer is B since the molecular mass of X would be equal to or greater than Y, and B can't be possible because Y is bigger.
Since Y is less than TWICE as heavy, you can't have X3Y because Y would be three times as heavy.

#### Trolly

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Hello

I'm doing a neutralization problem with a weak acid and a strong base. It gives you the molarity and volume of the weak acid and the molarity of base and asks you to find the volume of base. The answer key says that you use simple stoichiometry and moles of the base should equal moles of the acid because they are both monoprotic. But shouldn't there be less base added since it is a weak acid and will not dissociate completely? How come you do not have to use the Ka to figure out how many protons dissociate in the weak acid, and then calculate the volume of base relative to that?

Thanks!

#### BerkReviewTeach

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Hello

I'm doing a neutralization problem with a weak acid and a strong base. It gives you the molarity and volume of the weak acid and the molarity of base and asks you to find the volume of base. The answer key says that you use simple stoichiometry and moles of the base should equal moles of the acid because they are both monoprotic. But shouldn't there be less base added since it is a weak acid and will not dissociate completely? How come you do not have to use the Ka to figure out how many protons dissociate in the weak acid, and then calculate the volume of base relative to that?

Thanks!
Neutralization does not mean "to make pH = 7." It means to completely consume the reagent in solution. If you have a mole of acid, whether it's strong or weak, you require one mole of strong base to neutralize it.

#### TypeSH07

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How come in a reaction between NaNO2 and water the NO2- reacts with water and not the Na+??

#### RPedigo

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How come in a reaction between NaNO2 and water the NO2- reacts with water and not the Na+??
Let's say we have Na+ and NO2- in water. To simplify it, let's just say we have some H+ and some OH- in the water from the self-ionization of water.

If OH- and Na+ try to combine, it'll break right apart-- NaOH is a strong base and dissociates completely. There will be no NaOH in the water, it'll all be Na+ and OH- so there is no net reaction.

If H+ and NO2- try to combine, we get HNO2, a weak acid, nitrous acid. This won't all dissociate, and dissociates only slightly since it's a weak acid, so you will have HNO2 present in water just floating around all attached and stuff.

If it were NO3- instead of NO2-, then it wouldn't react since HNO3 is a strong acid, nitric acid, and would dissociate completely.