Question

Given: 50 ml of 0.2 M weak acid is titrated with 0.2M NaOH. The pka of the acid is 4.7

1. What is the pH at the equivalence point?

2. What is the ph when 60ml of NaOH is added? (Total Amt of NaOH is 60ml)

3. What is the pH when 10ml of NaOH is added?

4. What would be the approximate ph of a 0.1 M NaCl solution?

Answer 1

1 Equivalence point

AH + OH   H2O + A^-

	AH	OH	H2O	A-
ni	0.01	0.01	-	-
nrx	-0.01	-0.01		+0.01
nf	0	0		0.01

Then for opposite reaction

   H2O + A^-    AH + OH   

	A-	H2O	AH	OH
ni	0.01
nrx	-x		+x	+x
nf	0.01-x		x	x

Kb = 1E-14/10^-4.7= 5.01E-10

Kb = x²/0.01

x = (Kb x 0.01)^1/2 = 2.23E-6

[OH-] = 2.23E-6mol/100mL = 2.23E-5

pOH = -log 2.23E-6 = 4.65

pH = 9.35

2) pH when 60mL NaOH added

mol OH = 0.2/1000 x 60 = 0.012mol

nT = neq + nadd = 2.23E-6 + 0.012 = 0.012

[OH] = 0.012/160mL x1000mL = 0.075 M

pOH = 1.12

pH = 12.87

3) 10ml added OH

AH + OH   H2O + A^-

	AH	OH	H2O	A-
ni	0.01	0.002	-	-
nrx	-0.002	-0.002		+0.002
nf	0.008	0		0.01

Then for opposite reaction

   H2O + A^-    AH + OH   

	A-	H2O	AH	OH
ni	0.002		0.008
nrx	-x		+x	+x
nf	0.002		0.008	x

Kb = 1E-14/10^-4.7= 5.01E-10

Kb = 0.008x/0.002

x = Kb x 0.002/0.008 = 1.25E-10

[OH-] = 1.25E-10mol/110mL X 1000= 1.13E-9

pOH = -log 1.13E-9= 8.9

pH = 5.04