estion 3 of 8 General Chemistry 4th Edition Map df University Science Books pres
ID: 550384 • Letter: E
Question
estion 3 of 8 General Chemistry 4th Edition Map df University Science Books presented by Sapling Leaming Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.140 M HCIO(aq) with 0.140 M KOH(aq). The ionization constant for HCIO can be found here. Number (a) before addition of any KOH Number (b) after addition of 25.0 mL of KOH Number (c) after addition of 30.0 mL of KOH Number (d) after addition of 50.0 mL of KOH Number (e) after addition of 60.0 mL of KOH Hint O Previous Give Up & View Solution Check AnswerNext -el Exit about us careers privacy policy terms of uExplanation / Answer
a) HClO(aq) + H2O(l) <--------> H3O+(aq) + ClO-(aq)
Ka = [ H3O+ ][ClO-]/[HClO]=2.95×10^-8
x^2/(0.140 - x) = 2.95×10^-8
assume 0.140 - x = 0.140
x^2 = 4.13 × 10^-9
x = 6.42 × 10^-5
[ H3O+ ] = 6.42 ×10^-5
pH = - log(6.42×10^-5)
= 4.19
b) 25ml is half equivalene point
at half equivalence point, pH = pKa
pKa of HClO = 7.53
Therefore,
pH = 7.53
c) No of mole of HClO = (0.140mol/1000ml)×50ml = 0.0070
No of mole of KOH added = (0.140mol/1000ml)×30ml = 0.0042
OH + HClO -------> H2O + ClO-
0.0042mole of KOH react with 0.0042 mole of HClO to produce 0.0042 mole of ClO-
After addition
No of mole of HClO = 0.0070 - 0.0042 = 0.0028
No of mole of ClO- = 0.0042
total volume = 80ml
[HClO] = (0.0028mol/80ml)×1000ml = 0.0350M
[ClO-] = (0.0042mol/80ml) ×1000ml = 0.0525M
pH = pKa + log ( [ A- ]/[ HA]
= 7.53 + log ( 0.0525M/0.0350)
= 7.53 + 0.18
= 7.71
d) 50ml is equivalence point
at equivslence point
No of mole of ClO- formed = 0.0070
Volume = 100ml
[ ClO- ] =( 0.0070mol/100ml)×1000ml = 0.070M
ClO- (aq) + H2O(l) <------> HClO(aq) + OH- (aq)
Kb = [ HClO] [ OH- ]/[ClO- ] = 3.39×10^-7
x^2/0.070 = 3.39×10^-7
x = 1.54×10^-4
[OH- ] = 1.54×10^-4M
pOH = - log(1.54×10^-4)
= 3.81
pH + pOH = 14
pH = 14 - 3.81 = 10.19
e) Out of 60ml ,50ml of KOH is consumed for neutralization of HClO
excess KOH added =10ml
No of mole of KOH =( 0.140mol/1000ml)×10ml = 0.0014mol
Total volume = 110ml
[ OH- ] = (0.0014mol/110ml)×1000ml = 0.0127M
pOH = -log(0.0127)
=1.90
pH = 14 - 1.90
= 12.10
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