Question

To a 25.00 mL volumetric flask, a lab technician adds a 0.375 g sample of a weak monoprotic acid, HA , and dilutes to the mark with distilled water. The technician then titrates this weak acid solution with 0.0923 M KOH . She reaches the endpoint after adding 44.93 mL of the KOH solution.

After the technician adds 15.35 mL of the KOH solution, the pH of the mixture is 5.73. Determine the p?a of the weak acid.

Answer 1

let us consider the weak monoprolie and to be HA. let t' mole of MA be consumed, HA - HE) & A(-) Initial į 0 0 At equilibrium e(1-2) en la [HA] As is very less, Nowo, [at] = ca y log (A) = log (cm) pH = -log(cm) When 15:35 and of K014 soln is added, the pH of the mixture is 5.73 " N, S, & V2 S2 >> 25XS, - 15-35 x 0.0923 4 S 0.0567 (M) Me concentration of mono protic and = 0.0567 (M) Also pH = 5.73 - log (0.0567x) = 5.73 oy 0.0567 0 105.73

2 10 5.73 - 2 3.28x10-5 0.0567 2.98 %1005] We know, Kaca2 As a ús vory small compared to 1, inka a cm2 & Ka 2 0.0567 +(3:28)² x 10-10 sylka a 0.61410-10 log Ka a log (0.61 x 1070) log Ka a ~10.21 . Ipka a 10.21 ka of weak monoprotic add a 10.21)