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4 years ago

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What is the pH of a solution prepared by mixing 25.00 mL of 0.10 M CH3CO2H with 25.00 mL of 0.010 M CH3CO2Na? Assume that the vo

lume of the solutions are additive and that K a = 1.8 × 10-5 for CH3CO2H.

1 answer:

denpristay [2]4 years ago

4 0

Answer:

pH = 3.74

Explanation:

<u>Given:</u>

Initial volume of CH3COOH, V1 = 25.00 ml

Initial concentration of CH3COOH, M1 = 0.10 M

Initial volume of CH3COONa, V1 = 25.00 ml

Initial concentration of CH3COONa, M2 = 0.010 M

Ka (CH3COOH) = 1.8*10^-5

<u>To determine:</u>

pH of the solution

<u>Calculation:</u>

The given solution of CH3COOH/CH3COONa is in fact a buffer whose pH is given by the Henderson-Hasselbalch equation where:

$pH = pKa + log\frac{[A-]}{[HA]} ----(1)$

where A- = concentration of conjugate base = [CH3COONa]

HA = weak acid = [CH3COOH]

Step 1: Calculate the final concentration of CH3COONa

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.010 M

$M1V1 = M2V2\\\\M2 = \frac{M1V1}{V2} = \frac{0.010 M * 25.00 ml}{50.00ml} =0.005M$

<u>Step 2</u>: Calculate the final concentration of CH3COOH

V1 = 25.00 ml

V(final) = Total volume = 25.00 + 25.00 = 50.00 ml

M1 = 0.10 M

$M1V1 = M2V2\\\\M2 = \frac{M1V1}{V2} = \frac{0.10 M * 25.00 ml}{50.00ml} =0.05M$

<u>Step 3</u>: Calculate the pH

Based on equation (1)

$pH = pKa + log\frac{[CH3COONa]}{[CH3COOH]} ----(1)$

pKa = -log Ka = -log(1.8*10^-5) = 4.74

$pH = -logKa + log\frac{[CH3COONa]}{[CH3COOH]}$

$pH = 4.74 + log\frac{[0.005]}{[0.05]}$

pH = 3.74

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