During a titration of a weak base with a strong acid, you are slowing converting

Question

During a titration of a weak base with a strong acid, you are slowing converting molecules of the weak base into molecules of its conjugate add. For the hypothetical weak base, B we see the following: B (aq) + H_3O^+ (aq) rightarrow BH^+ (aq) + H_2O (l) In the problem below you will be adding some strong acid, but not enough to reach the endpoint of the titration. You are titrating 24.00 mL of a 1.50 M solution of C_6H_5NH_2 (aniline) with a strong acid. If you add 1.00 mL of a 1.00 M solution of hydrochloric acid, what is the final pH of the remaining weak base solution? The K_b for C_6H_5NH_2 (aniline) is 3.9 times 10^-10.

Explanation / Answer

V = 24 mL of aniline

M = 1.50 M

strong acid...

V = 1 mL f HCl, M = 1 M

find pH in weak solution:

mmol of base = MV = 24*1.5 = 36 mmol of base

mmol of strong acid = MV = 1*1 = 1 mmol of acid

then, after reaction

mmol of base left = 36-1 = 35

mmol of conjguate formed = 0 +1 = 1

this is abuffer since there is aniline and its conjguate C6H5NH3+

so

pOH = pKb + log(C6H5NH3+ / C6H5NH2)

pKb = -log(Kb)= -log(3.9*10^-10) = 9.4089

pOH = 9.4089+ log(C6H5NH3+ / C6H5NH2)

pOH = 9.4089+ log(1/35) = 7.864

pH = 14-pOH = 14-7.864 = 6.136 approx

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