Question #122344
9. What is the pH 500 mL of solution containing 0.0124 grams of Ca(OH)2? 10. The pH of a solution is 4.80. What is the concentration of hydroxide ions in this solution? 11. A solution in which [H^+] = 10^-8 M has a pH of _ and is_. 12. The pH of a 0.02 M of an unknown weak acid is 3.7. What is the pKa of this acid? 13. What is the approximate pH of a solution labeled 0.050 M HClO? 14. What is the pH of a solution labeled 0.30 M (CH3)3N ? 16. A 0.10 M solution of a weak acid, Hx, is 0.059% ionized. Evaluate Ka for the acid. 17. What is the percent ionization of an 1.2 M HF solution?
1
Expert's answer
2020-06-16T10:16:27-0400

9.

pH=log[H+]=14log[OH]pH = -log[H^+] = 14 - log[OH^-]

[OH]=c(OH)=n(OH)/V[OH^-] = c(OH^-) = n(OH^-)/V

[OH]=m(Ca(OH)2)M(Ca(OH)2V=0.012474.0930.5=0.000335[OH^-] = {m(Ca(OH)_2) \over M(Ca(OH)_2V} = {0.0124 \over 74.093*0.5} = 0.000335

pH=143.47=10.53pH = 14-3.47 = 10.53

10.

[OH]=10pOH=10(14pH)[OH^-] = 10^{-pOH} = 10^{-(14-pH)}

[OH]=10(144.80)=109.20=1.581010[OH^-] = 10^{-(14-4.80)} = 10^{-9.20} = 1.58*10^{-10}

11. 8 and is base

12.

[H+]=10pH=103.7=0.0002[H^+] = 10^{-pH} = 10^{-3.7} = 0.0002

ka=α2c=(0.0002)20.02=81010k_a = \alpha^2c = (0.0002)^2*0.02 = 8*10^{-10}

13.

pH100.050=0.89pH \approx 10^{-0.050} = 0.89

14.

pH100.3=140.50=13.50pH \approx 10^{-0.3} = 14 - 0.50 = 13.50

16.

ka=α2c=0.05920.10=0.00035k_a = \alpha^2c = 0.059^20.10 = 0.00035


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