Question

Calculate the ph of a solution that is made by mixing 50.0 ml of 0.14 m acetic acid and 50.0 ml of 0.21 m sodium acetate. express your answer to two decimal places.

Answer 1

After mixing 50.0 mL and 50.0 mL the molarities are cut in half to 
0.07 M acetic acid and 0.105 M sodium acetate 

the Ka for acetic acid = 1.8 x 10^-5

HA => H+ & A- 

1.8 x 10^-5 = [H+ ] [A-] / {HA] 

1.8 e-5 = [H+ ] [0.105] / [0.07] 

H+ = 1.2 x 10^-5 M 

pH = - log (1.2 x10^-5 M)

pH = 4.92

Answer 2

The pH of the resulting solution is 4.92.

From the information in the question, we mixed 50.0 mL solutions of acetic acid and sodium acetate hence the concentration of each solution was reduced to half.

• 0.07 M acetic acid
• 0.105 M sodium acetate

The Ka for acetic acid is 1.8 x 10^-5 hence;

HA ⇄H+ & A-

1.8 x 10^-5 = [H+ ] [A-] / {HA]

1.8 x 10^-5 = [H+ ] [0.105] / [0.07]

H+ = 1.2 x 10^-5 M

pH = - log (1.2 x10^-5 M)

pH = 4.92

Hence, the pH of the resulting solution is 4.92.

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