Important and quick: pKa and pH question

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orangeblue

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Given that the pKa for nitrous acid is 3.34, what is the pH of a 0.10 nitrous acid solution?

A. 1.00
B. 2.17
C. 3.24
D. 4.34

If it were a strong acid, pH = - log [H+] where [H+] = .10 b.c all of the acid dissolves.
Since nitrous acid is weak acid, HA <--> H+ + A, not all of HA is converted into H+ so the pH will be more basic than pH of 1. This rules out A.

Now, since pKa is when half of the acid has been dissolved. Meaning that HA <---> H+ and A- , where [HA] = H+ and [A-]

Initially , half of the acid hasn't been dissolved, as it's a weak acid and we haven't added any titrant (to cause the reaction to shift towards dissociation and increase [H+] ...currently very little H+ , so wouldn't the answer be D?

It's not. Tell me what's the missing gap in this reasoning.

Thanks!

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Calculation:

1. pKa = 3.34, coverted to Ka = ~ 5x10^-4 M
2. Ka expression: Ka = [H+][CB]/[HA] or Ka = [x^2]/[HA]
3. Solving for x: Ka*[HA] = x^2; 5x10^-4M * 0.10M = 5x10^-5 ===> sqrt(5x10^-5) = x; ~0.007 = x
4. -log(x) = pH so, pH = 2.15 (choice B is closest)

Now, since pKa is when half of the acid has been dissolved. Meaning that HA <---> H+ and A- , where [HA] = H+ and [A-]
Ka is an equilibrium expression for a given acid. It varies based on the strength of the given acid. pKa is just the -log of that value. I think you're confusing pKa and it's relevance to buffers, which is that at half equivalence point, pKa = pH. Totally different concept there.

For this question, just realize it's a weak base. It won't fully dissociate like a really strong acid would, hence ruling out A. This leaves choice B,C, and D. Only way to figure this out is by calculating. If you realize what the Ka expression tells you, like any equilibrium expression, then solving for the unknown variable, in this case [H+] concentration is fairly straightforward.

One thing to keep in mind though for acid dissociation, in particular weak acids/bases, is that because the amount that dissociates compared to the initial concentration is relatively small, we can neglect subtracting whatever [H+] ions were produced from the initial concentration of acid or base. In other words, for the purpose of calculating, you can neglect: [HA]-x and instead just use the initial concentration of 0.1. In a typical General Chemistry class, you would be required to use the quadratic equation, but for the MCAT, this estimation is okay.

From there, solving and doing math correctly will bring you to the correct answer.
 
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Calculation:

1. pKa = 3.34, coverted to Ka = ~ 5x10^-4 M
2. Ka expression: Ka = [H+][CB]/[HA] or Ka = [x^2]/[HA]
3. Solving for x: Ka*[HA] = x^2; 5x10^-4M * 0.10M = 5x10^-5 ===> sqrt(5x10^-5) = x; ~0.007 = x
4. -log(x) = pH so, pH = 2.15 (choice B is closest)

From there, solving and doing math correctly will bring you to the correct answer.

Without a calculator, step 1 alone would be tough...

Using Henderson Hasselbach you can guesstimate the answer much quicker. You know that strong acid dissociates completely and weak acids barely dissociates. As you said, a 0.1M strong acid would give you a pH of 1, so that rules out A.

pH = pKa + log A-/HA

We can also rule out D, because to get a pH value of +1 higher compared to the pKa from the HH equation requires the A-/HA ratio to be 10 (log of 10 = 1). For this much A- to be present, you have to titrate the acid with a base to take advantage of the le chatelier's principle, which isn't happening in this problem.

Now you have B and C left. There is two way to approach this. Either you realize now that in a weak acid, the dissociation is in the single digit percentage wise, which would mean the A-/HA ratio is in the 0.01-0.1 range, which means the log(A-/HA) is in the -1 to -2 range. This would give you a pH of 1.34-2.34 which is choice B.

The other way is to realize that if choice C were to be true:
0.1 = log(A-/HA)
10^0.1 = A-/HA
10th root of 10... it's going to be something close to 1 or a 50/50 ratio. Using the same logic for ruling out D, you realize that such a high dissociation ratio would make this a borderline strong acid, which you know is false for nitrous acid. Or that you are titrating this with a base, which is also false.
 
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Two approaches here:

1) Conceptual: HNO2 is a weak acid, so it does not fully dissociate, which means the pH cannot be as low as 1.0. A buffer, a mixture with equal parts HNO2 and NO2-, would have pH = pKa, which is 3.34 in this case. Because you have a solution rich in weak acid with very little weak base, the pH must be lower than the pKa by more than 1, given that it is not a buffer. The only answer between 1.00 and 2.34 is choice B.

2) Math: HNO2 is a weak acid, so the Berkeley Review Weak Acid Equation can (and should) be employed. We simply have to average 1.00 and 3.34, which is 2.17. Using the BR approach takes about 10 seconds.
 
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Can someone explain this question without a calculator
if HCLO4 has aka of 7.28, then the ph of a .10 M HCLO solution is which of the following?
a. 3.63
b. 4.13
c. 7.28
d. 8.26
 
I guess you meant that the pKa of HClO = 7.28? HClO4 is a strong acid, would dissociate completely.

Shortcut equation

pH = 1/2*pKa - 1/2 * log[HA]
pH = 0.5*7.28 - (1/2 * log(.1)
log(.1)= -1
pH = 3.64 - (-0.5)
pH = 4.14
 
can you explain why in a buffer solution, it says you need equal molars of base and acid , but if the base is stronger than the acid, then you need half?
 
I think that maybe what you're getting at is that if you have a weak acid and a weak base, then for your acidic species to be equal in molarity to your basic species, you need equal amounts of your weak acid and your weak base? Like, if you wanted to create a buffer of acetic acid, than you could mix together 10 mol of acetic acid with 10 mol of acetate. That makes sense.

What if you don't have any acetate lying around, but you still want to make this buffer? Well, than you could do this.

You could add 20 mol of acetic acid, and 10 mol of NaOH. NaOH as you know is a very strong salt that will dissolve completely into 10M Na+ and 10M OH-, and OH- is a very strong base that will be very happy to take any protons away from even weak acids such as acetic acid.

So when you add 10 mol of NaOH to your solution with 20 mol of acetic acid, those 10 mol of OH- are going to take all of the protons from 10 mols of your acetic acid. So it'll look like this:

20 mol acetic acid + 10 mol OH- --> 10 mol acetic acid + 10 mol acetate + 10 mol H2O

Make sense? To be clear, OH- is the limiting reagent here. If you added 15 mol of NaOH, then you'd get 15 mol acetate and only 5 mol of acetic acid. This is still a buffer, but it's a worse one than the one in which you have equal amounts of your acid and your conjugate base.
 
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