A)when 0.0 mL of NaOH is added
HClO dissociates as:
HClO -----> H+ + ClO-
6*10^-2 0 0
6*10^-2-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 ((2.9*10^-8)*6*10^-2) = 4.171*10^-5
since c is much greater than x, our assumption is correct
so, x = 4.171*10^-5 M
use:
pH = -log [H+]
= -log (4.171*10^-5)
= 4.3797
Answer: 4.38
B)when 12.5 mL of NaOH is added
Given:
M(HClO) = 0.06 M
V(HClO) = 50 mL
M(NaOH) = 0.12 M
V(NaOH) = 12.5 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.06 M * 50 mL = 3 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.12 M * 12.5 mL = 1.5 mmol
We have:
mol(HClO) = 3 mmol
mol(NaOH) = 1.5 mmol
1.5 mmol of both will react
excess HClO remaining = 1.5 mmol
Volume of Solution = 50 + 12.5 = 62.5 mL
[HClO] = 1.5 mmol/62.5 mL = 0.024M
[ClO-] = 1.5/62.5 = 0.024M
They form acidic buffer
acid is HClO
conjugate base is ClO-
Ka = 2.9*10^-8
pKa = - log (Ka)
= - log(2.9*10^-8)
= 7.538
use:
pH = pKa + log {[conjugate base]/[acid]}
= 7.538+ log {2.4*10^-2/2.4*10^-2}
= 7.538
Answer: 7.54
C)when 25.0 mL of NaOH is added
Given:
M(HClO) = 0.06 M
V(HClO) = 50 mL
M(NaOH) = 0.12 M
V(NaOH) = 25 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.06 M * 50 mL = 3 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.12 M * 25 mL = 3 mmol
We have:
mol(HClO) = 3 mmol
mol(NaOH) = 3 mmol
3 mmol of both will react to form ClO- and H2O
ClO- here is strong base
ClO- formed = 3 mmol
Volume of Solution = 50 + 25 = 75 mL
Kb of ClO- = Kw/Ka = 1*10^-14/2.9*10^-8 = 3.448*10^-7
concentration ofClO-,c = 3 mmol/75 mL = 0.04M
ClO- dissociates as
ClO- + H2O -----> HClO + OH-
0.04 0 0
0.04-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 ((3.448*10^-7)*4*10^-2) = 1.174*10^-4
since c is much greater than x, our assumption is correct
so, x = 1.174*10^-4 M
[OH-] = x = 1.174*10^-4 M
use:
pOH = -log [OH-]
= -log (1.174*10^-4)
= 3.9302
use:
PH = 14 - pOH
= 14 - 3.9302
= 10.0698
Answer: 10.07
D)when 40.0 mL of NaOH is added
Given:
M(HClO) = 0.06 M
V(HClO) = 50 mL
M(NaOH) = 0.12 M
V(NaOH) = 40 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.06 M * 50 mL = 3 mmol
mol(NaOH) = M(NaOH) * V(NaOH)
mol(NaOH) = 0.12 M * 40 mL = 4.8 mmol
We have:
mol(HClO) = 3 mmol
mol(NaOH) = 4.8 mmol
3 mmol of both will react
excess NaOH remaining = 1.8 mmol
Volume of Solution = 50 + 40 = 90 mL
[OH-] = 1.8 mmol/90 mL = 0.02 M
use:
pOH = -log [OH-]
= -log (2*10^-2)
= 1.699
use:
PH = 14 - pOH
= 14 - 1.699
= 12.301
Answer: 12.30
Show ALL your calculations and the respective dissociation equations for each step. Assignments without calculations or...
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