use:
pKa = -log Ka
7.53 = -log Ka
Ka = 2.951*10^-8
1)when 0.0 mL of NaOH is added
HClO dissociates as:
HClO -----> H+ + ClO-
0.18 0 0
0.18-x x x
Ka = [H+][ClO-]/[HClO]
Ka = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Ka = x*x/(c)
so, x = sqrt (Ka*c)
x = sqrt ((2.951*10^-8)*0.18) = 7.288*10^-5
since c is much greater than x, our assumption is correct
so, x = 7.288*10^-5 M
use:
pH = -log [H+]
= -log (7.288*10^-5)
= 4.1374
Answer: 4.14
2)when 11.4 mL of NaOH is added
Given:
M(HClO) = 0.18 M
V(HClO) = 66 mL
M(NaOH) = 0.15 M
V(NaOH) = 11.4 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.18 M * 66 mL = 11.88 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.15 M * 11.4 mL = 1.71 mmol
We have:
mol(HClO) = 11.88 mmol
mol(NaOH) = 1.71 mmol
1.71 mmol of both will react
excess HClO remaining = 10.17 mmol
Volume of Solution = 66 + 11.4 = 77.4 mL
[HClO] = 10.17 mmol/77.4 mL = 0.1314M
[ClO-] = 1.71/77.4 = 0.0221M
They form acidic buffer
acid is HClO
conjugate base is ClO-
Ka = 2.951*10^-8
pKa = - log (Ka)
= - log(2.951*10^-8)
= 7.53
use:
pH = pKa + log {[conjugate base]/[acid]}
= 7.53+ log {2.209*10^-2/0.1314}
= 6.756
Answer: 6.76
3)
find the volume of NaOH used to reach equivalence point
M(HClO)*V(HClO) =M(NaOH)*V(NaOH)
0.18 M *66.0 mL = 0.15M *V(NaOH)
V(NaOH) = 79.2 mL
Given:
M(HClO) = 0.18 M
V(HClO) = 66 mL
M(NaOH) = 0.15 M
V(NaOH) = 79.2 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.18 M * 66 mL = 11.88 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.15 M * 79.2 mL = 11.88 mmol
We have:
mol(HClO) = 11.88 mmol
mol(NaOH) = 11.88 mmol
11.88 mmol of both will react to form ClO- and H2O
ClO- here is strong base
ClO- formed = 11.88 mmol
Volume of Solution = 66 + 79.2 = 145.2 mL
Kb of ClO- = Kw/Ka = 1*10^-14/2.951*10^-8 = 3.389*10^-7
concentration ofClO-,c = 11.88 mmol/145.2 mL = 0.0818M
ClO- dissociates as
ClO- + H2O -----> HClO + OH-
0.0818 0 0
0.0818-x x x
Kb = [HClO][OH-]/[ClO-]
Kb = x*x/(c-x)
Assuming x can be ignored as compared to c
So, above expression becomes
Kb = x*x/(c)
so, x = sqrt (Kb*c)
x = sqrt ((3.389*10^-7)*8.182*10^-2) = 1.665*10^-4
since c is much greater than x, our assumption is correct
so, x = 1.665*10^-4 M
[OH-] = x = 1.665*10^-4 M
use:
pOH = -log [OH-]
= -log (1.665*10^-4)
= 3.7786
use:
PH = 14 - pOH
= 14 - 3.7786
= 10.2214
Answer: 10.22
4)when 82.2 mL of NaOH is added
Given:
M(HClO) = 0.18 M
V(HClO) = 66 mL
M(NaOH) = 0.15 M
V(NaOH) = 82.2 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.18 M * 66 mL = 11.88 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.15 M * 82.2 mL = 12.33 mmol
We have:
mol(HClO) = 11.88 mmol
mol(NaOH) = 12.33 mmol
11.88 mmol of both will react
excess NaOH remaining = 0.45 mmol
Volume of Solution = 66 + 82.2 = 148.2 mL
[OH-] = 0.45 mmol/148.2 mL = 0.003 M
use:
pOH = -log [OH-]
= -log (3.036*10^-3)
= 2.5176
use:
PH = 14 - pOH
= 14 - 2.5176
= 11.4824
Answer: 11.48
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