Question

How many grams of an acid, HA (FW 90.04, Ka=5.60x10^-2) should be mixed with 5.00 g of its conjugate base, A^- (FW 176.22) to give a pH of 1.10 when diluted to 200 mL?

Answer 1

Concentration of conjugate base = [A^-] = ( mass / molar mass) / volume of solution in L

                                                         = ( 5.00 g / 176.22(g/mol) ) / 0.200L

                                                         = 0.142 M

According to Henderson's equation,

pH = pKa + log([conjugate base]/[acid])

     = - log Ka + log([A^-] /[HA])

1.10 = - log(5.60x10^-2) + log([A^-] /[HA])

1.10 = 1.25 +log([A^-] /[HA])

log([A^-] /[HA]) = -0.152

[A^-] /[HA] = 10^-0.152

              = 0.705

0.142 /[HA] = 0.705

[HA] = 0.201 M

[HA] = ( mass of HA / molar mass of HA) / Volume of solution in L

0.201 = ( m / 90.04) / 0.200L

m = 3.624 g

Therefore the mass of acid added is 3.624 g