Calculate the pH of each of the solutions and the change in pH to 0.01 pH units
ID: 553325 • Letter: C
Question
Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.81-M HCl to 340. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.
a) water pH before mixing = Correct: Your answer is correct.
pH after mixing= Incorrect: Your answer is incorrect.
pH change = Incorrect: Your answer is incorrect.
b) 0.171 M C2H3O21-
pH before mixing =
pH after mixing=
pH change =
c) 0.171 M HC2H3O2
pH before mixing =
pH after mixing=
pH change =
d) a buffer solution that is 0.171 M in each C2H3O21- and HC2H3O2
pH before mixing =
pH after mixing=
pH change =
Explanation / Answer
pH calculations,
a) water
pH before mixing = 7
pH after mixing, pH = -log(2.81 M x 10 ml/350 ml) = 1.09
pH change = 1.09 - 7 = -5.91
b) 0.171 M C2H3O2-
hydrolysis of salt,
C2H3O2- + H2O <==> HC2H3O2 + OH-
let x amount hydrolyzed
Kb = 1 x 10^-14/1.8 x 10^-5 = [HC2H3O2][OH-]/[C2H3O2-]
5.55 x 10^-10 = x^2/0.171
x = [OH-] = 9.75 x 10^-6 M
pH before mixing,
pH = 14 - log(9.75 x 10^-6) = 8.99
after addition of HCl = 2.81 M x 10 ml = 28.1 mmol
initial moles of C2H3O2- = 0.171 M x 340 ml = 58.14 mmol
neutralization gives HC2H3O2 = 28.1 mmol
C2H3O2- remained = 58.14 - 28.1 = 30.04 mmol
pH after mixing,
pH = pKa + log(base/acid)
= 4.75 + log(30.04/28.1) = 4.78
pH change = 4.78 - 8.99 = -4.21
c) 0.171 M HC2H3O2
pH before mixing,
HC2H3O2 + H2O <==> C2H3O2- + H3O+
Ka = [C2H3O2-][H3O+]/[HC2H3O2]
1.8 x 10^-5 = x^2/0.171
x = [H3O+] = 1.75 x 10^-3 M
pH = -log[H3O+] = 2.75
pH after mixing
HCl is a strong acid, thus
pH = -log(2.81 M x 10 ml/350 ml) = 1.09
pH change = 1.09 - 2.75 = -1.66
d) buffer
pH before mixing
pH = pKa + log(base/acid)
= 4.75 + log(0.171/0.171) = 4.75
pH after mixing,
pH = 4.75 + log[(0.171 M x 340 ml - 2.81 M x 10 ml)/(0.171 M x 340 ml + 2.81 M x 10 ml)] = 4.29
pH change = 4.29 - 4.75 = -0.46
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