A buffer solution was prepared by dissolving 0.02 mole acetic acid & 0.01 mole sodium acetate in enough water to make 1.0 L of solution ( K a ( CH 3 COOH ) = 2 × 1 0 − 5 ) . If 5 × 1 0 − 4 mole Na

Question

A buffer solution was prepared by dissolving 0.02 mole acetic acid & 0.01 mole sodium acetate in enough water to make 1.0 L of solution ( K a ( CH 3 ​ COOH ) ​ = 2 × 1 0 − 5 ) . If 5 × 1 0 − 4 mole NaOH were added to 100 ml of the buffer. The resultant pH of the solution would be [given lo g 5 = 0.70 ]

TestHub · Accepted Answer

Initially

pH = pKa + log[Salt]/[Acid]

pH = 5 - log(2) = 4.7

[CH₃COOH] = 0.02M

[CH₃COONa] = 0.01M

In 100 mL; milimoles of CH₃COOH = 2

milimoles of CH₃COONa = 1

On addition of 0.5 milimole NaOH, it will consume 0.5 milimole of CH₃COOH and will produce 0.5 milimole of CH₃COONa.

Finally;

milimoles of CH₃COOH = 1.5

milimoles of CH₃COONa = 1.5

pH = pKa + log[Salt]/[Acid] = pKa = 5 - log(2) = 4.7

Chemistry - Ionic Equilibrium Question with Solution | TestHub

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