Question

Calculate how many milliliters of 0.663 M KOH should be added to 5.00 g of MOBS (see Table 8-2) to give a pH of 7.40. pKa=7.6. molecular weight=123.29

Answer 1

Given the mass of MOBS = 5.00 g

Molecular weight of MOBS = 123.29 g/mol

Hence moles of MOBS (weak acid - HA) = mass / molesular weight = 5.00 g / 123.29 gmol^-1  = 0.0406 mol

Given the concentration of KOH, M = 0.663 M

Let the volume of KOH taken = V L

Hence moles of KOH = MxV = 0.663xV mol

When the strong base KOH and weak acid HA are added A^-(aq) and water will be formed making the solution a buffer solution ( due to inter conversion of HA and A- ). The chemical reaction is

HA(aq) + OH-(aq) --> A^-(aq) + H2O

Initial mol: 0.0406 0.663xV 0   

eqm. mol: (0.0406 - 0.663xV) 0 0.663xV

Concentration of HA(aq) at eqm , [HA(aq)] =mol / Vt = (0.0406 - 0.663xV) / Vt

Concentration of A^-(aq) at eqm, [A^-(aq)] = mol / Vt = 0.663xV / Vt

Now applying Henderson equation

pH = pKa + log [salt] / [acid]

=> 7.40 = 7.6 + log[A^-(aq)] / [HA(aq)] = 7.6 + log [( 0.663xV / Vt) / (0.0406 - 0.663xV) / Vt]

=>  log [( 0.663xV) / (0.0406 - 0.663xV)] = - 0.20

=> ( 0.663xV) / (0.0406 - 0.663xV) = antilog ( - 0.20) = 10^-0.20  = 0.631

=>0.663xV = 0.0406x0.631 - 0.663x0.631V = 0.0256 - 0.418V

=> 0.663V - 0.418V = 0.0256

=> V = 0.104L = 104 mL (answer)