Question

Assume you dissolve 0.235g of the weak acid benzoic acid in enough water to make 1.00*102 mL of soultion and then titrate the solution with 0.108M NaOH. What is the pH at the equivalence point? When 1.55g of solid thallium(I) bromide is added to 1.00L of water, the salt dissolves to a small extent TlBr\left(s\right)\longleftrightarrow Tl^+\left(aq\right)+Br^-\left(aq\right). The thallium(I) and bromide ions in equilibrium with TlBr each have a concentration of 1.9*10-3 M. What is the value of pKsp for TlBr?

Answer 1

moles benzoic acid = 0.235 g / 122.12 g/mol=0.00192
At equivalence point all the acid will be neutralize by the base. So, moles NaOH required = 0.00192
volume of NaOH required= 0.00192 / 0.108 M=0.0178 L
total volume = 0.0178 L + 0.100 L = 0.118 L
moles benzoate formed = 0.00192
concentration benzoate = 0.00192 / 0.118 L=0.0163 M

C6H5COO- + H2O <=> C6H5COOH + OH-
Kb = Kw/Ka = 1.0 x 10^-14 / 6.3 x 10^-5 = 1.6 x 10^-10 M

1.6 x 10^-10 = x^2 / 0.0163-x
x = [OH-]= [C6H5COOH]= 1.61 x 10^-6 M
[H+]= 1.0 x 10^-14 / 1.61 x 10^-6=6.21 x 10^-9
pH = 8.21

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[TlBr] = 1.55g/284.29 gmol^-1 = 5.45 *10^-3 moles

Molarity of [TlBr] = 5.45 *10^-3 M

TlBr (s) <==> Tl (aq) + Br-(aq)

Ksp = [Tl+][Br-] = 5.45 *10^-3 * 5.45 *10^-3 = 2.97 *10^-5

pKsp = -log Ksp = 4.53