Question

A 41.00 mL mixture containing 0.0727 M Tl⁺ and 0.0727 M Hg₂²⁺ is titrated with 0.0867 M NaI resulting in the formation of the precipitates TlI and Hg₂I₂. Calculate pI^- at the following points in the titration. The Ksp values for TlI and Hg₂I₂ are 5.54 × 10^-8 and 2.9 × 10^-29, respectively.

a. 43.3 mL

b. 82.2 mL

c. the second equivalence point

d. 110.4 mL

Answer 1

To calculate the ionic point, the concentrations of the ions in solution must be known:

a) PI (TI2) = [TI +] * [I-]

PI (Hg2I2) = [Hg2 + 2] * [I-] 2

we calculate them by dilution ratio C1 * V1 = C2 * V2

V2 = 41 + 43.3 = 84.3 mL

[TI +] = [Hg2 + 2] = 0.0727 * (41 mL / 84.3 mL) = 0.035 M

[I-] = 0.0867 * (43.3 / 84.3) = 0.044 M

PI (TI2) = 0.035 * 0.044 = 1.54x10 ^ -3

PI (H2I2) = 0.035 * 0.044 ^ 2 = 6.78x10 ^ -5

b) V2 = 125.5 mL

[TI +] = [Hg2 + 2] = 0.0727 * (41 mL / 125.5 mL) = 0.024 M

[I-] = 0.0867 * (82.2 / 125.5) = 0.057 M

PI (TI2) = 0.024 * 0.057 = 1.36x10 ^ -3

PI (H2I2) = 0.024 * 0.057 ^ 2 = 7.80x10 ^ -5

c) V2 = 151.4 mL

[TI +] = [Hg2 + 2] = 0.0727 * (41 mL / 151.4 mL) = 0.020 M

[I-] = 0.0867 * (110.4 / 151.4) = 0.063 M

PI (TI2) = 0.020 * 0.063 = 1.26x10 ^ -3

PI (H2I2) = 0.020 * 0.063 ^ 2 = 7.94x10 ^ -5