Question

Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.120 M HClO(aq) with 0.120 M KOH(aq). The ionization constant for HClO can be found here. (a) before addition of any KOH (b) after addition of 25.0 mL of KOH (c) after addition of 40.0 mL of KOH (d) after addition of 50.0 mL of KOH (e) after addition of 60.0 mL of KOH

Answer 1

(a) Before adding any KOH

	HClO	H+	ClO-
inital	0.12	0	0
change	-x	+x	+x
equilibrium	0.12-x	+x	+x

K = [H+][ClO-]/[HClO]

4 *10^-8 = x*x/(0.12-x)

x is very small, so 0.12-x ~0.12 . Now solve the above equation for x.

x = 0.69 *10^-4

[H+] = 0.69 *10^-4

pH = -log [0.69 *10^-4] = 4.16

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(b) When KOH is added, it will react with the acid to form a buffer system (weak acid and its salt) . pH of a buffer can be calculated by using Hinderson hasselbalch equation.

pH = pka + log [salt/acid]

concentration of salt = concentration of base added

moles of base added = 0.12 * 25 mL /1000 = 0.003 moles. total volume = 50+ 25 = 75mL

Molarity = 0.003 moles *1000mL/75mL = 0.04 M

Moles of acid initially present = 0.12 *50/1000 mL = 0.006 moles . When a base is added to this acid solution 0.003 moles with react with base to form a salt. So, this amount has to be deducted from the initial moles present.

Molarity of the acid after addition = 0.003*1000mL/75 mL = 0.04M

pH = pka + log [CLO-/HClO]

= -log (4 *10^-8) + log [0.04/0.04]

= 7.39

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(c) Moles of base added = 0.12M * 40mL/1000mL = 0.0048 moles

Moles of salt formed = 0.0048 moles

Moles of acid present after the addition = 0.006-0.0048 = 1.2 *10^-3 moles

pH = pka + log [0.0048 moles/1.2 *10^-3]

= 7.39 +0.60 = 7.99

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(d) Moles of base added = 0.12 *50mL/1000mL = 0.006 moles

moles of acid = 0.006-0.006 = 0 . That is the all the acid has been neutralize by the base.

Concentraion of ClO- = 0.006 *1000mL/75mL = 0.08 M

ClO- will react with water to form HClO and OH-.

ClO- + H2O ---> HClO + OH-

	ClO-	H2O	HClO	OH-
inital	0.006	--	0	0
change	-x	--	+x	+x
equilibrium	0.006-x		0.006+x	0.006+x

Kb = kw/Ka = 1*10^-14/4*10^-8 = 0.25 *10^-6

Kb = [HClO][OH-]/[ClO-]

0.25 *10^-6 = x^2/0.006-x

x = 1.2X10^-4

[OH-] = 1.2X10^-4, pOH = log [1.2X10^-4]= 3.92 So, pH = 14-pOH = 10.08