Asked by nelson
1. What is the [K+] of a solution prepared by dissolving 0.140 g of potassium hydroxide in
sufficient pure water to prepare 250.0 ml of solution?
2. What is the pH of a 0.570 M solution of aniline? Kb = 7.4 × 10-10
sufficient pure water to prepare 250.0 ml of solution?
2. What is the pH of a 0.570 M solution of aniline? Kb = 7.4 × 10-10
Answers
Answered by
DrBob222
Convert 0.140 g KOH to mols. mols = grams/molar mass.
There is 1 mol K^+ per mol KOH.
Then [K^+] = mols/0.250 L = ??
Let's designate, for simplicity in typing, aniline as RNH2.
RNH2 + HOH ==> RNH3^+ + OH^-
Kb = (RNH3^+)(OH^-)/(RNH2) = 7.4 x 10^-10
(RNH3^+) = x
(OH^-) = x
(RNH2) = 0.570 - x
Substitute into Kb and solve for x = (OH^-), convert to (H^+) and solve for pH. Post your work if you get stuck.
Check my thinking.
There is 1 mol K^+ per mol KOH.
Then [K^+] = mols/0.250 L = ??
Let's designate, for simplicity in typing, aniline as RNH2.
RNH2 + HOH ==> RNH3^+ + OH^-
Kb = (RNH3^+)(OH^-)/(RNH2) = 7.4 x 10^-10
(RNH3^+) = x
(OH^-) = x
(RNH2) = 0.570 - x
Substitute into Kb and solve for x = (OH^-), convert to (H^+) and solve for pH. Post your work if you get stuck.
Check my thinking.
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