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Given: Ka(HF) = 6.3EE-04 A solution is prepared by adding 0.057 moles of calcium fluoride to...

Given: Ka(HF) = 6.3EE-04 A solution is prepared by adding 0.057 moles of calcium fluoride to a 150 mL volumetric flask and diluting to the calibration mark. What is the pH of this solution?

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Answer #1

Number of moles of CaF2(aq) = 0.057 mol

Volume of solution = 150 ml = 150 L / 1000 = 0.150 L

[CaF2] = 0.057 mol / 0.150 L = 0.38 mol/L = 0.38 M

In aqueous solution, [F^-(aq)] = 2 * [CaF2(aq)] = 2 * 0.38 M = 0.76 M

We know . Kb = Kw / Ka = 1.00 * 10^-14 / 6.3 * 10^-4 = 0.1587 * 10^-10

Consider a hydrolysis reaction of F^-(aq)

F^-(aq) + H2O (l) ----> OH^-(aq) + HF(aq)

Initial Conc. 0.76 0 0

Change in Conc. - x. +x +x

Final Conc. 0.76 - x . x. x

Kb = [OH^-][HF] / [F^-] = (x)(x) / (0.76 - x) = x^2 / 0.76

Since Kb is very small hence 0.76 - x ~ 0.76

x^2 = 0.76 * Kb = 0.76 * 0.1587 * 10^-10 = 0.1206 * 10^-10

x = √(0.1206 * 10^-10) = 0.3473 * 10^-5 = 3.473 * 10^-6

Since [OH^-] = x M hence [OH^-] = 3.473 * 10^-6 M

We know

pOH = - log[OH^-] = - log (3.473 * 10^-6) = - log 3.473 + 6 log 10 = - 0.54 + 6 * 1 = 5.46

We know pH = 14.00 - pOH = 14.00 - 5.46 = 8.54

pH = 8.54

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