Calculate the pH of the solution after 0.0470 L of LiOH is added. The expression

Question

Calculate the pH of the solution after 0.0470 L of LiOH is added. The expression for K_b for this reaction is the following. K_b = [HCN][OH^-]/[CN^-] Because the cyanide ion is a weak base, we can make the simplifying assumption that its equilibrium concentration is the same as the concentration we calculated in Step 9. From the balanced equation for the reaction of cyanide with water we see that the equilibrium concentrations of hydrocyanic acid and hydroxide are equal Therefore, we can solve for (OH^-). (Enter an unrounded value.) K_b = [HCN][OH^-]/[CN^-] [OH^-] = [HCN] K_b = [HCN][OH^-]/[CN^-] = [OH^-]^2/[CN^-] 1.63e - 05 = [OH^-]^2/0.600 [OH^-]^2 = 14.51 [OH^-] = Squareroot 14.51 3.8091 The pOH of the solution is the following. (Enter an unrounded value.) pOH = -log[OH^-] = -log(3.8091) .5808 We know that pH + pOH = 13.996, so we can calculate pH from pOH. pH + pOH = 13.996 pH + .5808 = 13.996 (Enter an unrounded value.) pH = 13.42 Since the (OH^-] was only only known to two significant figures, the pH must be rounded to two decimal places.

Explanation / Answer

SOLUTION:

The values like Kb are not given in question but are given in the answer you have attempted. So here we insert the values that you have given in your attempted answer.

Kb = [OH-]2 / [CN-]

1.63 X 10-0.5 = [OH-]2 / 0.060

[OH-]2 = 0.060 X 1.63 X 10-0.5 = 0.0309

[OH-] = 0.1758M

pOH = -log(0.17) = 7.5498

pH = 13.996 - 7.5498 = 6.45

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