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For 500.0 mL of a buffer solution that is 0.155 molL−1 in CH3COOH and 0.135 molL−1...

For 500.0 mL of a buffer solution that is 0.155 molL−1 in CH3COOH and 0.135 molL−1 in CH3COONa, calculate the initial pH and the final pH after adding 0.020 mol of HCl.

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Homework Answers

Answer #1

We know that,

pKa for CH3COOH = 4.76

Given,

[CH3COOH] = 0.155 mol / L = 0.155 M

[CH3COONa] = 0.135 M

1) Initial pH

pH of an acidic buffer = pKa + log (salt / Acid)

=> pH = pKa + log ([CH3COONa] / [CH3COOH])

=> pH = 4.76 + log (0.135 / 0.155) = 4.7

2) After adding HCl

Moles of HCl added = 0.02

Volume of solution = 500 mL = 0.5 L

=> Molarity of HCl = X = 0.02 / 0.5 = 0.04 M

pH after 'X' M of HCl = pKa + log ([CH3COONa] - X / [CH3COOH] + X)

=> pH = 4.76 + log (0.135 - 0.04 / 0.155 + 0.04) = 4.45

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