check the solution

[inaccensa]

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How much solid NaOH is required to neutralize 700
mL of 2 N H2SO4?

There are 2 equivalent of H+ in one mole of H2SO4,so 2 N simply means in
terms of molarity 1mole/L and for 700ml

1mole * 1L * 700ml = 0.7moles
1L 1000ml
the moles of NaOH will be 0.7 *40g/mol = 28 g,but since for every one mole of H2SO4, we need 2 moles of NaOH, the total amt will be 28g*2 =56gms

If the same q were asked for lets say Ca(OH)2, then you'd have 0.7 moles *100g/mole =70g

 

[Omni]

Is this a Kaplan question? I've never seen normality used in AAMC questions, but I believe what I have is right:


1 M of H2SO4 = 2 N of H+ ions.
So what you actually have is just 1 M of H2SO4.
So that translates to 2 M of H+ ions. Multiply that with 0.7 L and that gives you a total of 1.4 moles of total H+. To neutralize that, you need equal amount of NaOH.

so 1.4 mols of NaOH= 40g * 1.4 = 56 grams.
Had the acid been a 1 N, then you would have just needed 28 grams.
a 3 N acid would require you 84 grams of NaOH

 

[eric51]

Looks right to me Omni.

 

[Omni]

lol, hopefully I'll get a score like yours on my MCAT this friday to speak for that 😛

 

[inaccensa]

Is this a Kaplan question? I've never seen normality used in AAMC questions, but I believe what I have is right:


1 M of H2SO4 = 2 N of H+ ions.
So what you actually have is just 1 M of H2SO4.
So that translates to 2 M of H+ ions. Multiply that with 0.7 L and that gives you a total of 1.4 moles of total H+. To neutralize that, you need equal amount of NaOH.

so 1.4 mols of NaOH= 40g * 1.4 = 56 grams.
Had the acid been a 1 N, then you would have just needed 28 grams.
a 3 N acid would require you 84 grams of NaOH

Ok I did the problem based on a solution from kaplan.
This is solution for 2N HNO3
To solve this problem, you need to remember the definition of normality, and know how acids and bases interact to neutralize each other. Normality is defined as the number of equivalents per liter of solution. Here nitric acid has one equivalent, that is, it has one proton to donate pr mole. Thus, for this compound, normality equals molarity, and we have 2 moles in one liter, or 1.4 moles in 700 milliliters. Since acids and bases react in a 1 to 1 ratio of equivalents, in order to neutralize 1.4 moles of nitric acid, you must have an equal number of equivalents of the neutralizing substance
present. Since sodium hydroxide also has 1 equivalent per mole, we need the same number of moles of each. Therefore, we need 1.4 moles of sodium hydroxide. Sodium hydroxide has a molar weight of 40 grams, 1.4 moles will have a mass of 56 grams. This is answer C, which is the correct choice.

 

[inaccensa]

Is this a Kaplan question? I've never seen normality used in AAMC questions, but I believe what I have is right:


1 M of H2SO4 = 2 N of H+ ions.
So what you actually have is just 1 M of H2SO4.
So that translates to 2 M of H+ ions. Multiply that with 0.7 L and that gives you a total of 1.4 moles of total H+. To neutralize that, you need equal amount of NaOH.

so 1.4 mols of NaOH= 40g * 1.4 = 56 grams.
Had the acid been a 1 N, then you would have just needed 28 grams.
a 3 N acid would require you 84 grams of NaOH

It looks like though wt u and I did is pretty much the same, I multiplied the final ans with 2, since 1 mole of NaOH gives 1 N OH

 

[Omni]

I always just see normality as a different way of saying as the concentration of H+ ions.
For example, 1 M of H2SO4 is going to be 2N which is 2 M of H+

 

[eric51]

I always just see normality as a different way of saying as the concentration of H+ ions.
For example, 1 M of H2SO4 is going to be 2N which is 2 M of H+

That's exactly what it is 🙂
Well at least 99% of the time, that is what it's referring to.