15.) a.)In the laboratory, a general chemistry
student measured the pH of a 0.376 M aqueous
solution of nitrous acid to be
1.872.
Use the information she obtained to determine the Ka for
this acid.
Ka(experiment) =
b.)In the laboratory, a general chemistry
student measured the pH of a 0.376 M aqueous
solution of nitrous acid to be
1.870.
Use the information she obtained to determine the Ka for
this acid.
Ka(experiment) =
c.)In the laboratory, a general chemistry
student measured the pH of a 0.376 M aqueous
solution of hydrocyanic acid to be
4.928.
Use the information she obtained to determine the Ka for
this acid.
Ka(experiment) =
15)
a) [HNO2] = 0.376 M
pH = -log[H+] = 1.872
[H+] = 0.01343 M
[H+] = [A-] = 0.01343 M
[HNO2] at equilibrium = 0.376 - 0.01343 = 0.3626 M
Ka (experimental) = [H+][A-]/[HA]
= (0.01343)^2/(0.3626)
= 4.97 x 10^-4
---
b) [HNO2] = 0.376 M
pH = -log[H+] = 1.870
[H+] = 0.01350 M
[H+] = [A-] = 0.01350 M
[HNO2] at equilibrium = 0.376 - 0.01350 = 0.3625 M
Ka (experimental) = [H+][A-]/[HA]
= (0.01350)^2/(0.3625)
= 5.03 x 10^-4
--
c) [HCN] = 0.376 M
pH = -log[H+] = 4.928
[H+] = 1.18 x 10^-5 M
[H+] = [A-] = 1.18 x 10^-5 M
[HCN] at equilibrium = 0.376
Ka (experimental) = [H+][A-]/[HA]
= (1.18 x 10^-5)^2/(0.376)
= 3.70 x 10^-10
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