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