EK 1001 gen chem pg 87 number 816

[2010premed]

A 0.05 M solution of conjugate base of acis S would have a pH of approximately:

A: 9

50 mL of acid S are used, and 50mL of 0.1M NaOH are used to bring it to the eq point. at half eq pH = 6 and therefore pKa = 6

The explanation says that at the equivalence, only the conjugate base of S is left. However, this solution has been diluted by 2 by the titrant, so the equivalence point reflects the pH of a 0.05M solution of the conjugate base.

How do we know that the solution has been siluted by 2? Is this b/c NaPH converted double the amount of acid S to its conj base?

Also, how would I figure out the pH knowing the pKA of S?

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[TieuBachHo]

First, you need to compare them in term of [moles] ratio. However, there is a quicker way to solve this problem. Seeing that they have the same amount of solutions which are 50ml, and the acid is only half that of the base in term of concentration. Therefore, we know that this is a 2 to 1 ratio. In other words, you will need twice as much acid to titrate an equivalent of the base.

At half equivalent point, pH = pKa due to the fact that the conc. amount of acid equals to that of the base (If you don't know this, just look it up in the text book).

 

[MediCynical]

Just looked at it. Here's how I'd do it (long way compared to their explanation):

pKa of S = 6, so pKb = 8 and therefore Kb = 1E-8

Kb = [HA][OH-]/[A-] = x^2/.05 {was .05-x, but x is negligible}

Solve for x, which is the [OH-], which then allows you to figure out the [H+] via [OH-][H+] = 1E-14

Use [H+] to figure out pH