with steps please 3. A 20.00 mL aliquot of KCGH702(aq) was titrated with 0.1150
ID: 304070 • Letter: W
Question
with steps please
3. A 20.00 mL aliquot of KCGH702(aq) was titrated with 0.1150 M HCl(aq) using both an indicator and a pH meter. Kg for sorbic acid, HCGH702, is 1.7 x105. A total of 35.54 mL of 0.1150 M HCI(aq) was required to reach the equivalence point (20 pts). a) Calculate the initial molarity and pH of the potassium sorbate solution. Calculate the pH at the half-equivalence and equivalence points of the titration and suggest an appropriate indicator for titration. Explain your choice. b) ndicator Phenolphthalein Bromothymol blue 7.0 Methyl red Thymol blue Methyl violet k, 80Explanation / Answer
Titration
1. KC6H7O2 + HCl --> HC6H7O2 + KCl
a) moles HCl reacted = 0.1150 M x 35.54 ml = 4.0871 mmol
moles KC6H7O2 present = 4.0871 mmol
initial molarity KC6H7O2 = 4.0871 mmol/20 ml = 0.2044 M
KC6H7O2 --> K+ + C6H7O2-
C6H7O2- + H2O <==> HC6H7O2 + OH-
let x amount of C6H7O2- hydrolyzed
Kb = [HC6H7O2][OH-]/[C6H7O2-]
1 x 10^-14/1.7 x 10^-5 = x^2/0.2044
x = [OH-] = 1.1 x 10^-5 M
pOH = -log[OH-] = 4.96
pH = 14 - pOH = 9.04
b) pH at half-equivalence
at half-equivalence, half of KC6H7O2 has reacted with HCl
[KC6H7O2] remained = [HC6H7O2] formed
pH = pKa + log(KC6H7O2/HC6H7O2)
= pKa
= -log(1.7 x 10^-5)
= 4.77
pH at equivalence point
all of KC6H7O2 neutralized to form HC6H7O2
[HC6H7O2] formed = (0.2044 M x 20 ml)/55.54 ml = 0.074 M
HC6H7O2 <==> H+ + C6H7O2-
let x amount dissociated
Ka = [C6H7O2-][H+]/[HC6H7O2]
1.7 x 10^-5 = x^2/0.074
x = [H+] = 1.12 x 10^-3 M
pH = -log[H+] = 2.95
c) after HCl = 0.1150 M x 36.25 ml added = 4.17 mmol
excess [H+] from HCl = (4.17 - 4.0871) mmol/56.25 ml = 0.0015 M
pH = -log[H+] = -log(0.0015) = 2.82
d) plot below
data point
HCl (ml) added pH
0 9.04
17.77 4.77
35.54 2.95
36.25 2.82
Plot volume HCl added on x-axis and pH on y-axis
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