Calculate the pH of each of the following aqueous solutions. (Enter your answers to two decimal places.) (a) 10.0 mL deionized water WebAssign will check your answer for the correct number of significant figures. 2.72 Incorrect: Your answer is incorrect. (b) 10.0 mL deionized water plus 5.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures. (c) 10.0 mL deionized water plus 10.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures. (d) 10.0 mL deionized water plus 15.0 mL of 0.10 M NaOH WebAssign will check your answer for the correct number of significant figures.

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jufsanabriasa jufsanabriasa · Answer 1 · 2021-03-22T01:44:32+01:00

Answer:

a. pH = 7.0

b. pH = 12.52

c. pH = 12.70

d. pH = 12.78

Explanation:

a. Deionized water has the [H⁺] of pure water = 1x10⁻⁷ (Kw = 1x10⁻¹⁴ = [H⁺][OH⁻] - [H⁺] = [OH⁻ -)

pH = -log[H⁺] = 7

b. Moles NaOH = 5x10⁻³L * (0.10mol / L) = 5x10⁻⁴moles OH⁻ / 0.015L = 0.0333M = [OH⁻]

-Total volume = 10mL+5mL = 15mL = 0.015L

pOH = -log[OH⁻] = 1.48

pH = 14-pOH

pH = 12.52

c. Moles NaOH = 0.010L * (0.10mol / L) = 1x10⁻³moles OH⁻ / 0.020L = 0.0500M = [OH⁻]

-Total volume = 10mL+10mL = 20mL = 0.020L

pOH = -log[OH⁻] = 1.30

pH = 14-pOH

pH = 12.70

d. Moles NaOH = 0.015L * (0.10mol / L) = 1.5x10⁻³moles OH⁻ / 0.025L = 0.060M = [OH⁻]

-Total volume = 10mL+15mL = 25mL = 0.025L

pOH = -log[OH⁻] = 1.22

pH = 14-pOH

pH = 12.78