Question

Consider the following data for questions A-F: Two solutions of an unknown slightly soluble salt, A(OH),, were allowed to equilibrate- one at 25°C and the other at 80*C. A 15.00 mL-aliquot of each solution is titrated with 0.200 M HCI. 5.47 mL of the acid is required to reach the endpoint of the titration at 25 °C, while 67.04 mL mL are required for the 80 °C solution.

Answer 1

K_sp at 80 ^oC = 0.357

Explanation

At 80 ^oC, volume of HCl required = 67.04 mL

concentration of HCl = 0.200 M

moles HCl required = (concentration of HCl) * (volume of HCl required)

moles HCl required = (0.200 M) * (67.04 mL)

moles HCl required = 13.408 mmol

moles OH^- present = moles HCl required

moles OH^- present = 13.408 mmol

[OH^-] = (moles OH^- present) / (volume of aliquot)

[OH^-] = (13.408 mmol) / (15.00 mL)

[OH^-] = 0.894 M

[A²⁺] = [OH^-]/2

[A²⁺] = (0.894 M) / 2

[A²⁺] = 0.447 M

K_sp = [A²⁺][OH^-]²

K_sp = (0.447 M) * (0.894 M)²

K_sp = 0.357