Calculate the pH of a solution that contains 1.3 M HF and 1.3 M HOC6H5.?

Calculate the pH of a solution that contains 1.3 M HF and 1.3 M HOC6H5.?

pKa of HF = 3.17

Ka = 10^-3.7 = 6.761 * 10^-4

Ka = [H+] * [F-] ÷ [HF] = 6.761 * 10^-4

HF → H+ + F-

let x = number moles of HF that dissociate in 1 liter of solution.

[H+] = x

[F-] = x

[HF] = 1.3 – x

x * x ÷ (1.3 – x) = 6.761 * 10^-4

x^2 = (1.3 – x) * 6.761 * 10^-4

x^2 = 8.7893 * 10^-4 – 6.761 * 10^-4 *x

x^2 + 6.761 * 10^-4 *x – 8.7893 * 10^-4 = 0

x = 0.0256

[H+] = 0.0256

pKa of HOC6H5 = 9.95

Ka = 10^-9.95 = 1.122 * 10^-10

Ka = [H+] * [OC6H5-] ÷ [HOC6H5] = 1.122 * 10^-10

HOC6H5 → H+ + OC6H5-

let x = number moles of HOC6H5 that dissociate in 1 liter of solution.

[H+] = x

[OC6H5-] = x

[HOC6H5] = 1.3 – x

x * x ÷ (1.3 – x) = 1.122 * 10^-10

x^2 = (1.3 – x) * 1.122 * 10^-10

x^2 = 1.4586 * 10^-10 – 1.122 * 10^-10 * x

x^2 + 1.122 * 10^-10 *x – 1.4586 * 10^-10 = 0

x = 1.2077 * 10^-5

[H+] from HOC6H5 = 1.2077 * 10^-5

[H+] from HF= 0.0256 = 2.56 * 10^-2

Since 1.2077 * 10^-5 is so much smaller than 2.56 * 10^-2, we will neglect 1.2077 * 10^-5

Total H+ = 2.56 * 10^-2

pH = -log (2.56 * 10^-2) = 1.59

To determine the final concentration of OC6H5- ion at equilibrium, we use the Ka equation.

[H+] = 2.56 * 10^-2

[OC6H5-] = x

[HOC6H5] = 1.3 – 1.2077 * 10^-5

Since 1.2077 * 10^-5 is so much smaller than 1.3 we can neglect 1.2077 * 10^-5.

[HOC6H5] = 1.3

1.122 * 10^-10 = 2.56 * 10^-2 * x ÷ 1.3

1.3 * 1.122 * 10^-10 ÷ 2.56 * 10^-2 = x

x = 5.7 * 10^-9 = [OC6H5-]