Question

Calculate the amount of ${\left( {N{H_4}} \right)_2}S{O_4}$ which must be added to $500{\text{ ml}}$ of $0.2{\text{ M N}}{{\text{H}}_3}$ to yield a solution of $pH{\text{ = 9}}{\text{.35 , p}}{{\text{K}}_{b{\text{ }}\left( {N{H_4}OH} \right)}}{\text{ = 4}}{\text{.74}}$ .

Answer 1

Hint: By using the value of $pH$ and $p{K_b}$ of $N{H_3}$ , find out the numbers of moles of $N{H_4}OH$ present in the solution .Then assuming the numbers of moles of $N{H_4}^ + $ which is present in solution , calculate the moles of $\left( {N{H_4}} \right)S{O_4}$ .
Formula used:
$pOH{\text{ = p}}{{\text{K}}_b}{\text{ - log }}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$

Complete answer:
Since the $pH$ of given solution will be more than seven so its basic in nature .Therefore we have to do calculation with $pOH$.Hence we will find $pOH$ firstly ,
As we know ,
$pOH{\text{ + pH = 14}}$
$pH{\text{ = 9}}{\text{.35 }}\left( {given} \right)$
Therefore , $pOH{\text{ = 14 - 9}}{\text{.35}}$
$pOH{\text{ = 4}}{\text{.65}}$
Now we have $pOH{\text{ = 4}}{\text{.65}}$ and $p{K_b}{\text{ = 4}}{\text{.74}}$ . Also we have a relation between them as,
$pOH{\text{ = p}}{{\text{K}}_b}{\text{ - log }}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$
Now put values here and find out the logarithmic ratio of salt and base. Therefore,
$\log {\text{ }}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}{\text{ = p}}{{\text{K}}_b}{\text{ - pOH}}$
Substituting the given values,
$\log {\text{ }}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}{\text{ = 4}}{\text{.74 - 4}}{\text{.65}}$
$\log {\text{ }}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}{\text{ = 0}}{\text{.09}}$ , here salt is $N{H_4}OH$ and base is $N{H_4}^ + $ .
Therefore taking anti log both sides we get the ratio as ,
$\dfrac{{\left[ {N{H_4}OH} \right]}}{{\left[ {N{H_4}^ + } \right]}}{\text{ = 1}}{\text{.23}}$
Since the molarity of the solution is given as $0.2{\text{ M}}$ . Therefore ,
$\left[ {N{H_4}^ + } \right]{\text{ = }}\dfrac{{0.2}}{{1.23}}{\text{ = 0}}{\text{.1626 M}}$
Hence we can say that ${\text{ 0}}{\text{.1626 M}}$ of ${\left( {N{H_4}} \right)_2}S{O_4}$ is required for ${\text{ 0}}{\text{.2 M }}$ of $N{H_4}OH$ . Therefore for ${\text{ 0}}{\text{.5 L or 500 ml }}$ of $N{H_4}OH$ , the number of moles required is :
$0.1626{\text{ M }} \times {\text{ 0}}{\text{.5 L = 0}}{\text{.081 moles}}$
The molecular weight of ${\left( {N{H_4}} \right)_2}S{O_4}$ is :
$\left( {14{\text{ + 1 }} \times {\text{ 4}}} \right){\text{ }} \times {\text{ 2 + 32 + 16 }} \times {\text{ 4}}$
$132{\text{ g}}$
Therefore the weight required is moles multiply with molecular weight as:
${\text{0}}{\text{.018 }} \times {\text{ }}132{\text{ g }}$
$10{\text{ g}}$
Hence $10{\text{ g}}$ of ${\left( {N{H_4}} \right)_2}S{O_4}$ is required .

Note:
Since the value of $p{K_b}$ is given, we use that formula specifically. We can calculate the millimoles too if volume is taken in ml , since we use in litre we get moles to millimoles .While calculating the value of antilog make use of an anti log table. The theoretical molecular weight can be different with practical molecular weight but it is negotiable.