The pH of 0.050 M solution of benzoate, NaC_6H_5CO_2, is measured to be pH = 8.4

Question

The pH of 0.050 M solution of benzoate, NaC_6H_5CO_2, is measured to be pH = 8.40. a. Calculate the concentration of hydronium ion [H_3O^+] using the pH. b. Calculate the concentration of Hydroxide ion [OH^-] in this solution. c. Calculate the K_b for sodium benzoate, C_6H_5CO_2^- (aq) using the equilibrium expression in 1(b). If the actual values for K_a of benzoic acid is 6.5 times 10^-5, calculate actual value of K_b for benzoate ion. If not all of the sodium benzoate solid dissolves in solution Q. #3, before the pH is measured, what effect would it have on pH measured in Q # 5. And finally what effect would it have on calculation of K_b value in Q #5(c)? Higher, lower, or no effect. Explain a. pH? b. K_b?

Explanation / Answer

F)

pH = 8.40

[H3O+] = 10-pH

[H3O+] = 10-8.40

[H3O+] = 3.98 x 10-9 M

[OH-] = 1.0 x 10-14 / [H3O+]

[OH-] = 1.0 x 10-14 / 3.98 x 10-9

[OH-] = 2.5 x 10-6 M

pH = 1/2 [pKw + pKa + logC]

8.40 = 1/2 [14 + pKa + log 0.05]

16.8 = 14 + pKa - 1.30

2.8 = pKa - 1.30

pKa = 4.1

pKb = 14 - 4.1

pKb = 9.9

Kb = 10-pKb

Kb = 10-9.9

Kb = 1.26 x 10-10

G) Ka = 6.5 x 10-5

Kb = Kw /Ka

Kb = 1.0 x 10-14 / 6.5 x 10-5

Kb = 1.54 x 10-10

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