Consider the titration of a 25.0 mL solution of 0.155 M CH 3 COOH ( K a = 1.8 x

Question

Consider the titration of a 25.0 mL solution of 0.155 M CH3COOH (Ka = 1.8 x 10-5 ) with 0.250 M NaOH. What is the pH of the solution after the addition of 10.0 mL of the titrant? Please show all steps needed to solve this!

a. 2.45                    b. 2.96                    c. 3.17                    d. 4.05                    e. 5.00

For the titration, what is the pH of the sodium acetate solution produced at the endpoint of the titration? Please provide steps here as well!

a. 6.45                    b. 7.05                    c. 7.33                    d. 8.05                    e. 8.86

Explanation / Answer

*

Initial moles of acid = 0.155 * 25.0 / 1000 = 0.00388 mol

moles of base added = 0.250 * 10.0 / 1000 = 0.0025 mol

Final moles of acid = 0.00388 - 0.0025 = 0.00138 mol

final mole sof salt fomrd= 0.0025 mol

pH = pKa + Log[salt]/[acid]

pH = 4.74 + log(0.0025/0.00138) = 5.00

(D)

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