Asked by Anonymous
Calculate the pH of a solution obtained by mixing of 100 mL of a 10-3 M and 50 mL of a 10-2 M solution of CH3COONa.
Answers
Answered by
DrBob222
CH3COOH = HAc.
Calculate M of the solutions. millimols of each.
mmols = mL x M = 100 mL x 10^-3 = 0.1
mmols = 50 x 10^-2 = 0.5
Total = 0.6
M = mmols/mL = 0.6/150 mL = approx 0.004
.....Ac^- + HOH --> HAc + OH^-
I...0.004............0.....0
C....-x..............x.....x
E..0.004-x...........x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.004-x). Solve for x = (OH^-) and convert to pH.
Calculate M of the solutions. millimols of each.
mmols = mL x M = 100 mL x 10^-3 = 0.1
mmols = 50 x 10^-2 = 0.5
Total = 0.6
M = mmols/mL = 0.6/150 mL = approx 0.004
.....Ac^- + HOH --> HAc + OH^-
I...0.004............0.....0
C....-x..............x.....x
E..0.004-x...........x.....x
Kb for Ac^- = (Kw/Ka for HAc) = (x)(x)/(0.004-x). Solve for x = (OH^-) and convert to pH.
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