# Problem: A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solution, the pH was 3.8.Determine the acid dissociation constant for this weak acid at 25°C.A. 5.81x10-4B. 1.85x10-4C. 1.05x10-9D. 1.58x10-4E. 5.01x10-9

###### FREE Expert Solution

We’re being asked to determine the acid dissociation constant (Ka) for a weak monoprotic acid

Recall that at the equivalence point of a titration:

Also, recall that moles = molarity × volume

This means:

$\overline{){\left(\mathbf{MV}\right)}_{{\mathbf{acid}}}{\mathbf{=}}{\left(\mathbf{MV}\right)}_{{\mathbf{base}}}}$

The weak acid is monoprotic so it ha only one H+ to donate and one equivalence point:

HA H+ + A-        Ka                  Equivalence Point

(MA)acid = (MV)base

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###### Problem Details

A 0.70 g sample of a weak monoprotic acid was dissolved in 150 mL of water and titrated with a 0.055 M solution of KOH at 25°C. At the half-equivalence point, after addition of 25.5 mL of the KOH solution, the pH was 3.8.

Determine the acid dissociation constant for this weak acid at 25°C.

A. 5.81x10-4

B. 1.85x10-4

C. 1.05x10-9

D. 1.58x10-4

E. 5.01x10-9