Solution X | 100 mL of 0.10 M NaOH(aq) is mixed with 100 mL of 0.10 M HBr(aq) Solution Y | 100 mL of — Acids and Bases Chemistry Question
Problem Context
Solution X | 100 mL of 0.10 M NaOH(aq) is mixed with 100 mL of 0.10 M HBr(aq)
Solution Y | 100 mL of 0.10 M NaBr(aq) is mixed with 100 mL of 0.10 M HBr(aq)
Solution Z | 100 mL of 0.10 M HC2H3O2(aq) is mixed with 100 mL of 0.10 M NaC2H3O2(aq)
A student prepares three solutions, X, Y, and Z, as described in the table above. The values of Ka for the acidic species in the solutions are given in the table below.
Species | Ka
HBr(aq) | >>1 (very large)
HC2H3O2(aq) | 1.8×10^-5
Model Answer
Lowest pH: Y < Z < X: Highest pH
Solution Y is a strong acid solution with a very low pH. Solution Z is a buffer solution with pKa = 4.74 = pH. Solution X is a neutral solution created from equimolar amounts of a strong acid and a strong base that react in a 1:1 ratio.
1 point is earned for the correct ordering. 1 point is earned for a valid explanation of the ranking.
Model Answer
The pH of the solution increases. The addition of water will decrease [H+]; therefore, the pH will increase.
1 point is earned for the correct choice and a valid justification.
Model Answer
Solution Z is a buffer system (composed of a weak acid and its conjugate base), whereas solution Y is not a buffer.
1 point is earned for a valid explanation.