Question

A buffer solution made up of $BOH$ and $BCl$ of total molarity 0.29 M has $pH=9.6$ and $K_b=1.8\times {10}^{-5}$. Concentration of salt and base respectively is:

• A. 0.09 M and 0.2 M
• B. 0.2 M and 0.09 M
• C. 0.1 M and 0.19 M
• D. 0.19 M and 0.1 M

Answer 1

The correct option is A 0.09 M and 0.2 M

$pH+pOH=14$

$\therefore pOH=14-9.6=4.4$

$forbasicbuffer=pOH=p{K}_b+\log \frac{\left[salt\right]}{\left[Base\right]}$

$4.4=-\log (1.8\times {10}^{-5})+\log \frac{\left[salt\right]}{\left[Base\right]}$

$4.4=4.74+\log \frac{\left[salt\right]}{\left[Base\right]}$

$-0.34=\log \frac{\left[salt\right]}{\left[Base\right]}$

$\left[Base\right]=2.18\left[salt\right]$

Now,

Total molarity $=\left[salt\right]+\left[Base\right]=0.29M$

$\left[salt\right]3.18=0.29M$

$salt=0.09M$

$\therefore Base=0.29-0.09=0.2M$