Question #c8443
1 Answer
(a) before addition of any KOH. pH = 4.05
(b) after addition of 25.0 mL of KOH. pH = 7.4
(c) after addition of 35.0 mL of KOH. pH = 7.03
(d) after addition of 50.0 mL of KOH. pH = 10.7
(e) after addition of 60.0 mL of KOH. pH = 12.3
Explanation:
The reaction between HClO and KOH is:
HClO(aq) + KOH(aq) -> KClO(aq) + H2O(l)
We see that 1 mole HClO reacts with 1 mol KOH. Therefore, at every stage of titration, we can calculate the number of moles of acid reacting with base, and the pH of the solution is determined by the excess acid or base left over. At the equivalence point, however, the neutralization is complete, and the pH of the solution will depend on the extent of the hydrolysis of the salt formed, which is KClO.
(a) No KOH has been added. This is a weak acid calculation.
HClO (aq) + H2O(l) ↔ H3O+(aq) + ClO-(aq)
Initial (M): 0.200 0 0
Change (M): -x +x +x
Equilibrium (M): 0.200 - x x x
Assuming x<<0.200
pH = 4.05
(b) The number of moles of HClO originally present in 50.0 mL of solution is:
0.2M * 0.050L = 0.010
The number of moles of KOH in 25.0 mL is: 0.2M * 0.025L = 0.0050
We work with moles at this point because when two solutions are mixed, the solution volume increases. As the solution volume increases, molarity will change, but the number of moles will remain the same. The changes in number of moles are summarized.
HClO(aq) + KOH(aq) -> KClO(aq) + H2O(l)
Initial (mol): 1.00 x 10-2 0.50 x 10-2 0
Change (mol): -0.50 x 10-2 -0.50 x 10-2 +0.50 x 10-2
Final (mol): 0.50 x 10-2 0 0.50 x 10-2
At this stage, we have a buffer system made up of HClO and ClO- (from the salt, KClO). We use the Henderson-Hasselbalch equation to calculate the pH.
Ka of HClO = 4.0 x 10-8 pKa = 7.4
pH = pKa + log [salt] / [acid]
= pKa + log [KClO] / [HClO]
= 7.4 + log [ 0.05 mol] / [0.05 mol] = 7.4 – 0
pH = 7.4
(c) 10.0 x 10-3 mole of HClO reacts with 7.00 x 10-3 mole KOH to produce 7.00 x 10-3 mole of KClO.
The number of moles of KOH in 35.0 mL is: 0.2M * 0.035 L = 0.007 mol
The changes in number of moles are summarized.
HClO(aq) + KOH(aq) -> KClO(aq) + H2O(l)
Initial (mol): 10.0 x 10-3 7.00 x 10-3 0
Change (mol): -7.00 x 10-3 -7.00 x 10-3 +7.00 x 10-3
Final (mol): 3.0 x 10-3 0.0 x 10-3 7.00 x 10-3
At this stage, we have a buffer system made up of HClO and ClO- (from the salt, KClO). We use the Henderson-Hasselbalch equation to calculate the pH.
pH = pKa + log [salt] / [acid]
= pKa + log [KClO] / [HClO]
= 7.4 + log [ 0.007 mol] / [0.003 mol]
= 7.4 – 0.370
pH = 7.03
(d) We have reached the equivalence point of the titration. 1.00 x 10-2 mole of HClO reacts with 1.00 x 10-2 mole KOH to produce 1.00 x 10-2 mole of KClO. The only major species present in solution at the equivalence point is the salt, KClO, which contains the conjugate base, ClO-. Let’s calculate the molarity of ClO-. The volume of the solution is now 100mL.
= (50.0 ml x 0.2M) / 100 ml = 0.100M
Kb of ClO ion = Kw/ Ka of HClO
= (1x10-14) / (4.0 x 10-8)
= 2.50 x 10^-6
We set up the hydrolysis of ClO-, which is a weak base.
ClO-(aq) + H2O(l) ↔ HClO(aq) + OH-(aq)
Initial (M): 0.100 0 0
Change (M) -x +x +x
Equilibrium (M): 0.100 – x x x
x = 0.50 x 10-3 M = [OH-]
pOH = 3.30 ; pH= 14- pOH ; pH = 10.7
(e) We have passed the equivalence point of the titration. The excess strong base, KOH, will determine the pH at this point. The moles of KOH in 60.0 mL are:
0.2M * 0.060 L = 0.012 mol
The changes in number of moles are summarized.
HClO(aq) + KOH(aq) -> KClO(aq) + H2O(l)
Initial (mol): 10.0 x 10-3 12.00 x 10-3 0
Change (mol): -10.0 x 10-3 -10.00 x 10-3 10.0 x 10-3
Final (mol): 0 2.00 x 10-3 10.0 x 10-3
Let’s calculate the molarity of the KOH in solution.
The volume of the solution is now 110.0 mL = 0.110 L.
=(0.2M x 60.0 ml – 0.2M x 50.0 ml) / (60.0 ml + 50 ml) = 0.0181M
KOH is a strong base. The pOH is: pOH = -log(0.0181) = 1.74
pH = 12.3
