Acid-Base titration question

[DenTony11235]

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So we start with 50 mL of acid S, as pictured. It's at pH 4, has a pKa of 6, and the titration reaches it's equivalence pt. at 50 mL with .1M NaOH.

Question: What is the concentration of H+ at the beginning of the titration?

Using M(A)V(A)=M(B)V(B) at the equivalence point gives us M(A)=.1M.

I dont' understand why: pH= -log[H+] gives me 4=-log[.0001], with .0001M being [H+]. What the hell?

 

[DenTony11235]

Also: What is the approximate percent dissociation of acid S at the beginning of the titration?

I use pH= pka +log(CB/A) so 4=6+log(CB/A) which gives me .01 as the ratio of CB:A, or 1% dissociation. Why is this not true?

 

[Czarcasm]

So we start with 50 mL of acid S, as pictured. It's at pH 4, has a pKa of 6, and the titration reaches it's equivalence pt. at 50 mL with .1M NaOH.

Question: What is the concentration of H+ at the beginning of the titration?

Using M(A)V(A)=M(B)V(B) at the equivalence point gives us M(A)=.1M.

I dont' understand why: pH= -log[H+] gives me 4=-log[.0001], with .0001M being [H+]. What the hell?

Careful here. You're forgetting that you're starting with a weak acid. As you calculated, the concentration of HA is 0.1M. However, because this is a weak acid, only some of that will dissociate. Hence why we're at a pH of 4 instead of a pH of 1.

The second question is asking you the % dissociation. You have everything you need to know. What is % dissociaton of HA? Well, it's [H+]/[HA] x 100. A pH of 4 would indicate that 1x10^-4 H+ dissociated in solution. So: 1x10^-4 M H+ / 1x10^-1 M HA = 1x10^-3 x 100 = 0.1% dissociation.

 

[DenTony11235]

Great explanation. Thanks.