Question

7.(6 pts) Calculate the pH, [OH], and pOHl of a 0.0035 M HCl aqueous solution: 8. (6 pts) What are the [Hs0'1, (OH], and poH in a solution with a pH of 3.82? 9. (6 pts) What are the [H:0'], [OH], and pH in a solution with a pOH of 11.75? pH = pOH= , [OH'] = POH ' [OH'] = 10. (6 pts) Calculate the [H30, [OH1, and pH a 0.040 M Ba(OH)2 solution. [OH-11- pH=

Answer 1

7) 0.0035M HCl

HCl -----> H⁺ + Cl^-

[H⁺] = 0.0035M

K_w = 10^-14 = [H⁺][OH^-]

[OH^-] = 10^-14 / 0.0035 = 2.86 * 10^-12M

pH = -log[H⁺] = -log(0.0035) = 2.45

pOH = -log[OH^-] = -log(2.86 * 10^-12) = 11.54

8) pH = 3.82

[H⁺] = 10^-3.82 = 1.5 * 10^-4 M

[OH^-] = 10^-14/ 1.5 * 10^-4 = 6.67 * 10^-11M

pOH = -log(6.67 * 10^-11) = 10.17

9) pOH = 11.75

[OH^-] = 10^-11.75 = 1.77*10^-12M

[H⁺] = 10^-14 / 1.77*10^-12 = 5.6*10^-3 M

pH = -log(5.6*10^-3) = 2.25

10) Ba(OH)₂ ------> Ba²⁺ + 2OH^-

[OH^-] = 0.080M

[H⁺] = 10^-14 / 0.080 = 1.25*10^-13M

pH = -log(1.25*10^-13) = 12.90