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

Question

During a titration of a weak acid with a strong base, you are slowing converting molecules of the weak acid into molecules of its conjugate base. For the hypothetical weak acid, HA we see the following HA (aq) + OH' (aq) A. (aq) + H2O (l) Early in the titration, this is like a limiting reagent problem where the weak acid is present in excess. Each molecule of strong base converts one molecule of HA into one molecule of the conjugate base, A, so that the [HA] decreases and the [A] increases. As these two species (HA and A) return to equilibrium the pH is controlled by the common ion effect (just like in the common ion effect problems we have a lready worked). General Method: In the problem below you will be adding some strong base, but not enough to reach the endpoint of the titration. Use the solution stoichiometry to find out how many moles of the weak acid remain after the reaction with the strong base. Use this and the total volume to get the new weak acid and A concentrations. Then set it up like a common ion effect problem using the Ka, [HA] and [A] which was created by the strong base. 1.20 mL of the buret as the titrant 16.00 mL of original 0.90 sample In this problem you are titrating 16.00 mL of a 0.90 M solution of HCOOH with a strong base. If you add 1.20 mL of a 0.96 M solution of sodium hydroxide, what is the final pH of the acid solution? The Ka for HCOOH is 1.8 X 10 -4 pH37

Explanation / Answer

a)

mmol of acid = MV = 0.9*16 = 14.4

mmol of base = MV = 1.2*0.96 = 1.152

after reaction

mmol of acid left = 14.4 -1.152 = 13.248

mmol of HCOO- left = 1.152

apply buffer equation

pH = pKa + log(HCOO-/HCOOH)

pka = -log(1.8*10^-4) = 3.744

pH = 3.744+ log(1.152/13.248)

pH = 2.683

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