Titration question

[golgiapparatus88]

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What is the pH after 70mL of 0.20M HCl has been added to 50mL of 0.60M H3CNH2? Pka of H3CNH2 is 10.58.

TBR explains it in conceptual terms which is fine but I'm a math person. Unfortunately, I don't know how to solve this with the HH equation. I thought you do .07L x .2mol/L HCl = .014mol HCl and .05mL x .6m/L =.03mol H3CNH2. pH = 10.58 + log (.03/.014) = 10.91.

TBR says the answer is 10.65 which is close but not the answer. What am I doing wrong!! 😕

 

[pmahesh107]

You forgot to subtract the amounts. [acid] = .014 mol HCl / .12 L = .117 M. Your concentration of base with the new volume is .25 M. Subtract the acid to find the base left: .25 - .117 = .133 M. Now use the Henderson-Hasselbach equation: pH = 10.58 + log(.133/.117) you get about 10.64

 

[golgiapparatus88]

OH! Thank you!

 

[nUgnoY]

Hey there,

The way you did this question isn't the right way in my opinion. You might get a close answer, but through wrong math. Here's my thought process

1. Amines are weak bases. HCL is a strong acid. I am titrating a weak base by a strong acid.

2. I know I need to use something like the HH. But HH is for titrating a weak acid by strong base. I'll use similar steps to derive the counterpart equation for titrating a weak base by a strong acid. Skipping my derivation steps, eq is:
pOH = pKb + log ([HA]/[A-])*

3. Solve the eq. HA is the FINAL amount of conjugate acid and A- is FINAL amount of remaining base after complete reaction.

4. I have the pOH. pH = 14 - pOH

I get 10.638, which is slightly more exact but the numbers were probably rounded during the math in BR.

What you actually did was that you put

pH = pKa (of the BASE, not acid) + log (WeakBase/Strong Acid)

The pKa given is the pKa of the amine (RNH2) to be deprotonated to an amine anion (RNH-). What you want is the pK, with K being the equilibrium constant of amine (RNH2) going to aminium (RNH3+). Basically, you made an inappropriate substitution.

You put the concentration or the mole of WEAK BASE when you needed conjugate acid of weak base on the numerator on the ratio.

You put the concentration or the mole of STRONG ACID when you needed Weak Base on the denominator on the ratio.

My bottom line is that you might want to review titration and buffer and try some derivations to understand the concept better. If you're a math person, you should be well off after couple hours!

*The A- stands for base and HA for conjugate acid. Just as a conceptual double check, K = P/R. I'm making c. acid from base. So K=P/R, which is acid/base. So it makes sense (although during derivation there's actually 2 reciprocalization which cancels out so it's not directly this.

 

[Rabolisk]

The pKa given is not the amine being deprotonated. It's the pKa of the amine being protonated, or the protonated amine losing the extra proton.

You can get the answer by using pH = pKa + log [A-]/[HA], as long as you keep the acid-base pair consistent.

 

[nUgnoY]

The pKa given is not the amine being deprotonated. It's the pKa of the amine being protonated, or the protonated amine losing the extra proton.

You can get the answer by using pH = pKa + log [A-]/[HA], as long as you keep the acid-base pair consistent.

Actually this is very important. Amine should NOT have a pKa even close to 10.

Ammonium will have pKa around 10 because it's a weak base. But Amine losing a proton is sort of similar to carbon losing a proton to become a carbanion. A little easier, but not practical. In fact, pKa of ammonia (NH3) is 33.

So if you can write that pKa of amine is 10 and just move on, you should review acids and bases a little as well (because you should be able to have an idea whether a statement has sense or not; like if you'd see 1+1=3 you will immediate know something is wrong.) It'll help you to develop that sense for the MCAT

 

[Rabolisk]

The question assumes that when they say pKa of an amine, you will understand that strictly they mean the pKa of an ammonium ion. No one in this topic was confused about that.

 

[nUgnoY]

Oh I see. So MCAT or practice questions have actually said that pKa of an amine is 11 and implied assumption?

Interesting, that's definitely something to keep in mind. Thanks!

 

[BerkReviewTeach]

What is the pH after 70mL of 0.20M HCl has been added to 50mL of 0.60M H3CNH2? Pka of H3CNH2 is 10.58.

In case anyone is working through this in the future, the question is actually the following:

• What is the pH after 70mL of 0.20M HCl has been added to 50mL of 0.60M H3CNH2? H3CNH2 has a pKb of 3.42.

This means that the pKa of H3CNH3+ is 10.58.

You mix 0.014 moles HCl and 0.30 moles H3CNH2, which results in a solution with 0.016 moles H3CNH2 leftover and 0.014 moles H3CNH3+ formed. The HH equation would give you:

pH = 10.58 + log (0.016/0.014) = 10.58 + log 8/7 = 10.58 + log 8 - log 7 = 10.58 + 0.90 - 0.83 = 10.65.