Calculate the pH of each of the solutions and the change in pH to 0.01 pH units
ID: 553023 • 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 2.64-M HCl to 690. 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 =
pH after mixing=
pH change =
b) 0.152 M C2H3O21-
pH before mixing =
pH after mixing=
pH change =
c) 0.152 M HC2H3O2
pH before mixing =
pH after mixing=
pH change =
d) a buffer solution that is 0.152 M in each C2H3O21- and HC2H3O2
pH before mixing =
pH after mixing=
pH change =
Explanation / Answer
a) pH of water before addition =7
10.0 mL of 2.64-M HCl to 690. mL
total volume = 700mL
new concentration of HCl = initial concentration X initial volume / total volume = 2.64 X 10 / 700 = 0.0377
pH = -log[H+] = -log 0.0377 = 1.423
ph change = 7- 1.423 = 5.577
b) b) 0.152 M C2H3O2-
initial pH = ?
the acetate ion will hydrolyze as
CH3COO- + H2O --> CH3COOH + OH-
Initial 0.152 0 0
Change -x +x +x
Equilibrium 0.152-x x x
Kb = Kw /Ka = [CH3COOH] [ OH-] / [CH3COO-] = 10-14 / 1.8 X 10-5 = 5.5 X 10-10 = x2 / (0.152-x)
x <<1
Kb = 5.5 X 10-10 = x2 / (0.152
x = 9.14 X 10-6M = [OH-]
pOH = -log[OH-] = 5.04
pH = 14-5.04 = 8.96
after addition of HCl it will form some acetic acid
moles of acetic acid formed = Moles of HCl added = Molarity X volume in litres = 2.64 X 10 / 1000 = 0.0264
moles of acetate ion present initially = 0.152 X 690/1000 = 0.105 moles
moles of acetate left = Moles of acetate - Moles of acetic acid = 0.105 - 0.0264 = 0.0786
this will form buffer, pka of acetic acid is 4.74
pH = pKa + log [salt] / [acid]
pH = 4.74 + log [0.0786 / 0.0264] = 5.21
pH change = 8.96 - 5.21 = 3.75
c) 0.152 M HC2H3O2
pH before mixing will be due to acetic acid
[H+] = [Ka x concentration of acetic acid]1/2 = (1.8 X 10-5 x 0.0152)1/2 = 5.23 X 10-4 M
Moles of [H+] = Molarity X volume in litres = 5.23 X 10^-4 X 690 / 1000 = 0.000361
pH = -log 5.23 X 10-4 = 3.28
After addition of HCl the mole sof H+ added = Moles of HCl added = Molarity X volume in litres = 2.64 X 10 / 1000 = 0.0264
total moles of H+ = 0.000361 + 0.0264
[H+] = Moles / volume in litres = 0.0268 / 0.7 = 0.038
pH = -log 0.038 = 1.42
pH change = 3.28-1.42 = 1.86
d) a buffer solution that is 0.152 M in each C2H3O21- and HC2H3O2
Initial pH of buffer with acetate and acetic acid in same concentration = pKa of acetic acid = 4.74
after addition of HCl it will react with acetate ion to form acetic acid
moles of acetic acid formed = Moles of HCl added = Molarity X volume in litres = 2.64 X 10 / 1000 = 0.0264
moles of acetate ion present initially = 0.152 X 690/1000 = 0.105 moles
moles of acetic acid present initially = 0.152 X 690/1000 = 0.105 moles
Moles of acetic acid after addition of HCl =0.105 + 0.0264 = 0.1314
Moles of acetate after addition of HCl = 0.105 - 0.0264 =0.0786
pH = pKa + log [salt] / [acid]
pH = 4.74 + log [0.0786] / [0.1314] = 4.52
ph change = 4.74-4.52 = 0.22
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