Question

# $$100$$ mL of $$0.02\ M$$ benzoic acid ($$pKa = 4.20$$) is titrated using $$0.02\ M$$ $$NaOH$$. $$pH$$ after $$50$$ mL and $$100$$ mL of $$NaOH$$ have been added are:

A
3.50,7
B
4.2,7
C
4.2,8.1
D
4.2,8.25

Solution

## The correct option is D $$4.2, 7$$Given $$100ml$$ of $$0.02M$$ $${ C }_{ 6 }{ H }_{ 5 }COOH$$ with $$p{ K }_{ a }=4.2$$So, $${ K }_{ a }={ 10 }^{ -4.2 }=6.31\times { 10 }^{ -5 }$$It is titrated using $$0.02M$$ $$NaOH$$.$$(i)$$ After addition of $$50ml$$ of $$NaOH$$:       Number of moles of $${ C }_{ 6 }{ H }_{ 5 }COOH=100\times 0.02=2mmol$$      Number of moles of $$NaOH=50\times0.02=1mmol$$Here after the addition of $$50ml$$ of $$NaOH$$,we get basic Buffer solution  $${ C }_{ 6 }{ H }_{ 5 }COONa$$ and $${ C }_{ 6 }{ H }_{ 5 }COOH$$.Number of moles of $${ C }_{ 6 }{ H }_{ 5 }COOH=2-1=1mmol$$Number of moles of $${ C }_{ 6 }{ H }_{ 5 }COONa=1mmol$$$$pH=p{ K }_{ a }+log\ \dfrac {\text {salt}}{\text{acid}}$$ $$= 4.2 + log\dfrac{1}{1}$$$$pH = 4.2$$$$(ii)$$ After addition of $$100ml$$ of $$NaOH$$:       Number of moles of $${ C }_{ 6 }{ H }_{ 5 }COOH=100\times 0.02=2mmol$$      Number of moles of $$NaOH=100\times0.02=2mmol$$So, Number of moles of $${ C }_{ 6 }{ H }_{ 5 }COONa=2mmol$$Therefore the given point is equivalent point. The salt $${ C }_{ 6 }{ H }_{ 5 }COONa$$ is weak base, so we will find basic ionization constant.Now,$${ K }_{ b }=\frac { { K }_{ w } }{ { K }_{ a } } =\frac { { 10 }^{ -14 } }{ 6.31\times { 10 }^{ -5 } } =1.58\times { 10 }^{ -10 }$$Concentration of $$\left[ { C }_{ 6 }{ H }_{ 5 }COONa \right] =\frac { Number\quad of\quad moles\quad of\quad { C }_{ 6 }{ H }_{ 5 }COONa }{ Total\quad volume\quad of\quad the\quad solution } =\frac { 2 }{ 200 } =0.01$$$$\left[ O{ H }^{ - } \right] =\sqrt { { K }_{ b }\times \left[ { C }_{ 6 }{ H }_{ 5 }COONa \right] } =\sqrt { (1.58\times { 10 }^{ -10 })\times 0.01 } =1.257\times { 10 }^{ -7 }M$$$${ pOH }=-\log { \left[ O{ H }^{ - } \right] } =-\log { (1.257\times { 10 }^{ -7 }) } =6.9$$$${ pH }=14-{ pOH }=14-6.9 \approx 7.0$$Chemistry

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