Question

A) 300.0g of calcium hydroxide is placed in a 1L volumetric flask and the flask was filled to the mark with water at 25°C. There is obviously a lot of solid in the bottom of the flask. Ksp = 1.3 x 10-6 for calcium hydroxide. The contents of the flask are transferred quantitatively to a 2.00 L volumetric. 100. ml of 1.00 M HCL is added and the flask filled to the mark with water. There is still observable solid calcium hydroxide in the bottom of the new flask. What is the pH of this solution?

B)300.0g of calcium hydroxide is placed in a 1L volumetric flask and the flask was filled to the mark with water at 25°C. There is obviously a lot of solid in the bottom of the flask. Ksp = 1.3 x 10-6 for calcium hydroxide.

What is the pH of this solution?

Answer 1

300g of Ca(OH)_2 = 4.05 moles in 1L marked water Ca(OH)_2 rightleftharpoons Ca^2+ + 2OH^- x in moles/Litre Solubility product, Ksp = x middot (2x)^2 = 4x^3 = 1.3 times 10^-6 rightwardsdoublearrow x^3 = 0.325 times 10^-6 rightwardsdoublearrow x = 6.875 times 10^-3 moles/litre Therefore [OH^-] = 2x = 13.75 times 10^-3 moles/litre P^OH = -log[OH] = 1.87 rightwardsdoublearrow P^H = 14 - 1.87 = 12.13 ----Answer (B) *However, it was transferred to a^dot 2L^dot contain and added 100mL 1M Hcl So final volume = 1.1 L Therefore [OH^-] = 13.75 times 10^-3/1.1 moles/litre = 12.5 moles/litre Then, 100 mL of 1M HCl has 1100 mL has 0.091M HCl so concentration of acid, [H^+] = 0.0827 moles/litre Final concentration of [H^+] = (0.0827 - 13.75 times 10^-3) moles/Litre = 0.06895 Final P^H = -log[H^+] = -log(0.06895) = 1.16 ----Answer (A)