Calculate the pH for each case in the titration of 50.0 mL of 0.230 M HClO(aq)0.230 M HClO(aq) with 0.230 M KOH(aq).0.230 M KOH(aq). Use the ionization constant 4.0×10–8 for HClO.
What is the pH after the addition of 50.0 mL KOH?
Given:
M(HClO) = 0.23 M
V(HClO) = 50 mL
M(KOH) = 0.23 M
V(KOH) = 50 mL
mol(HClO) = M(HClO) * V(HClO)
mol(HClO) = 0.23 M * 50 mL = 11.5 mmol
mol(KOH) = M(KOH) * V(KOH)
mol(KOH) = 0.23 M * 50 mL = 11.5 mmol
We have:
mol(HClO) = 11.5 mmol
mol(KOH) = 11.5 mmol
11.5 mmol of both will react to form ClO- and H2O
ClO- here is strong base
ClO- formed = 11.5 mmol
Volume of Solution = 50 + 50 = 100 mL
Kb of ClO- = Kw/Ka = 1*10^-14/4*10^-8 = 2.5*10^-7
concentration ofClO-,c = 11.5 mmol/100 mL = 0.115M
ClO- dissociates as
ClO- + H2O
-----> HClO + OH-
0.115
0 0
0.115-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 ((2.5*10^-7)*0.115) = 1.696*10^-4
since c is much greater than x, our assumption is correct
so, x = 1.696*10^-4 M
[OH-] = x = 1.696*10^-4 M
use:
pOH = -log [OH-]
= -log (1.696*10^-4)
= 3.7707
use:
PH = 14 - pOH
= 14 - 3.7707
= 10.2293
Answer: 10.23
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