Question

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

Answer 1

Answer:

pH = 3.74

Explanation:

Given:

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

To determine:

pH of the solution

Calculation:

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

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

Step 2: 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

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

Step 3: 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

Answer 2

The addition of complexes to the solution changes the final concentration. The pH of the solution with the mixing of the two different solutions is 3.74.

What is the pH?

The pH has been the hydrogen ion concentration in the solution. It can be given with the acid dissociation ability of the compound, or the ability of a compound to release hydrogen ions.

The addition of 25 ml solutions resulted in the final volume of 50 ml. The final concentration of the solutions is given as:

$\rm Initial\;concentration\;\times\;Initial\;volume=Final\;concentration\;\times\;Final\;Volume$

The Final concentration of $\rm CH_3COONa$ salt is:

$\rm 0.01\;M\;\times\;25\;mL=50\;mL\;\times\;[CH_3COONa]\\\\CH_3COONa=\dfrac{0.01\;M\;\times\;25\;mL}{50\;mL}\\\\ CH_3COONa=0.005\;M$

The final concentration of $\rm CH_3COOH$ acid is :

$\rm 0.1\;M\;\times\;25\;mL=50\;mL\;\times\;[CH_3COOH]\\\\CH_3COOH=\dfrac{0.1\;M\;\times\;25\;mL}{50\;mL}\\\\ CH_3COOH=0.05\;M$

The pH of the solution is given as:

$\rm pH=log\;Ka\;+\;log\;\dfrac{salt}{acid}$

Substituting the values in the equation:

$\rm pH=-log\;1.08\;\times\;10^-^5\;+\;log\;\dfrac{0.005}{0.05} \\pH=4.74\;+\;(-1)\\pH=3.74$

The pH of the solution of sodium acetate and acetic acid is 3.74.

Learn more about the acid dissociation constant, here:

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