Calculate the pH of a solution by mixing 40.0 mL of a 0.02 M HCl solution with 200.0 mL of 0.20 M HCN solution. Assume volumes to be additive.Ka for HCN=1.0x10^-10

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sanjeetmanna | College Teacher | (Level 3) Assistant Educator

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HCl is a strong acid and HCN is a weak acid.

Given

Volume of HCl = 40ml = 0.04L

Molarity of HCl = 0.02M

Volume of HCN = 200ml = 0.2L

Molarity of HCN = 0.2M

First we have to find individual moles.

Moles = Molarity * volume.

Moles of HCl = 0.02 * 0.04 = 0.0008

Moles of HCN = 0.2 * 0.2 = 0.04

Total volume = 0.04 + 0.2 = 0.24

Total moles = 0.0008 + 0.04 = 0.0408

Total concentration = moles / volume = 0.0408/0.24 = 0.17

Ka = (x^2)/(0.17-x)

1.0x10^-10 = (x^2)/(0.17-x)

1.0x10^-10 = (x^2)/(0.17)

X^2 = (1.0x10^-10)(0.17)

X = 4.1 * 10^-6

Ph = -log(H+)

Ph = -log(4.1 * 10^-6)

Ph = 5.3.

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