Question

A buffer solution made up of BOH and BCl of total molarity $0.29{\text{ M}}$ has $pH{\text{ = 9}}{\text{.6}}$ and ${K_b}{\text{ = 1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}$ . Concentration of salt and base respectively is :
A.$0.09{\text{ M}}$ and ${\text{0}}{\text{.2 M}}$
B.${\text{0}}{\text{.2 M}}$ and $0.09{\text{ M}}$
C.${\text{0}}{\text{.1 M}}$ and $0.19{\text{ M}}$
D.$0.19{\text{ M}}$ and ${\text{0}}{\text{.1 M}}$

Answer 1

We can find the concentration of base and salt individually by finding the $pOH$ of the given buffer solution. Also the total molarity of solution is the sum of molarity of both salts and base. By using this molarity we can find their concentrations too.
Formula used:
$\left( {a.} \right)$ ${\text{pOH = }}p{K_b}{\text{ + log }}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$
$\left( {b.} \right)$ Total Molarity $= {\text{ }}\left[ {salt} \right]{\text{ + }}\left[ {base} \right]$

Complete answer:
It is a buffer solution of BOH and BCl having molarity of $0.29{\text{ M}}$. Here the $pH$ is given .Since we are given with ${K_b}$ and we need to find the concentration of base , so we have to find $pOH$ first.
To find $pOH$ of the given buffer solution :
We know that the sum of $pOH$ and $pH$ for an ideal solution is equal to $14$ .
Therefore ,
$pOH{\text{ + pH = 14}}$
$pOH{\text{ = 14 - pH}}$
Here we are given that : pH of the buffer solution is ${\text{9}}{\text{.6}}$.
Therefore :
$pOH{\text{ = 14 - 9}}{\text{.6}}$
$pOH{\text{ = 4}}{\text{.4}}$
Now we know that $pOH$ can be written in terms of ${K_b}$ and concentration of base and salt.
For basic buffer we have:
$pOH{\text{ = p}}{{\text{K}}_b}{\text{ + log}}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$
Here we have :
$pOH{\text{ = 4}}{\text{.4}}$
${K_b}{\text{ = 1}}{\text{.8 }} \times {\text{ 1}}{{\text{0}}^{ - 5}}$
Therefore on putting values we have :
$4.4{\text{ = - log}}\left( {1.8{\text{ }} \times {\text{ 1}}{{\text{0}}^{ - 5}}} \right){\text{ + log}}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$
$4.4{\text{ = 4}}{\text{.74 + log}}\dfrac{{\left[ {salt} \right]}}{{\left[ {base} \right]}}$
Taking antilog both sides we get:
$\left[ {base} \right]{\text{ = 2}}{\text{.18 }}\left[ {salt} \right]$
Now we know that the total molarity of solution is the sum of concentrations of both salt and base.
So we can write as :
Total Molarity ${\text{ = }}\left[ {salt} \right]{\text{ + }}\left[ {base} \right]$ $= {\text{ 0}}{\text{.29 M}}$
Using above result substitute $\left[ {base} \right]$
$\left[ {salt} \right]{\text{ 2}}{\text{.18 = 0}}{\text{.29 M}}$
$\left[ {salt} \right]{\text{ = 0}}{\text{.09 M}}$
Therefore
$\left[ {base} \right]{\text{ = 0}}{\text{. 29 M - 0}}{\text{.09 M}}$
$\left[ {base} \right]{\text{ = 0}}{\text{.2 M}}$
Hence the concentrations of both salt and base are thus calculated.
Thus the correct answer is option A.

Note:
Always keep a note that while writing formula make sure it is valid .If not first convert the parameters according to the formula required. For finding log use base ${\text{10}}$. Remember the log formulas for finding log values. Double check the log calculations.