Question

Calculate the weight in mg of HCl added to 100 ml of 0.1 N BOH to have its pH = 6.6 and $K_b=6.25\times {10}^{-8};$ Antilog $(-7.4)=3.98\times {10}^{-8}.$ (Assume there is no change in volume on addition of $HCl$)

• A. 300
• B. 100
• C. 425
• D. 223

Answer 1

The correct option is D 223

Given, $\left[BOH\right]=0.1N,pH=6.6,K_b=6.25\times {10}^{-8},V=100ml$
Let the mass of HCl added be x mg
$BOH+HCl\rightleftharpoons BCl+H_{2}O\text{Millimoles}10\frac{x}{36.5}00ateqb10-\frac{x}{36.5}0\frac{x}{36.5}$
This constitutes basic buffer.
For basic buffer, the $\left[O{H}^{-}\right]$ is given by Henderson equation
$pOH=-log{K}_b+log\frac{\left[BCl\right]}{\left[BOH\right]}$
$\left[O{H}^{-}\right]=K_b\frac{\left[BOH\right]}{\left[BCl\right]}..........\left(1\right)$
we know, $pH+pOH=14$
$6.6+pOH=14$
$pOH=7.4$
​​​​​​​$-log\left[O{H}^{-}\right]=7.4$
​​​​​​​$log\left[O{H}^{-}\right]=-7.4$
​​​​​​​$\left[O{H}^{-}\right]=3.98\times {10}^{-8}$
now substitue values in equation (1),
$3.98\times {10}^{-8}=6.25\times {10}^{-8}\times \left(\frac{365-x}{36.5}\right)\left(\frac{36.5}{x}\right)$
$\left(\frac{365-x}{36.5}\right)\left(\frac{36.5}{x}\right)=0.6368$
$\left(\frac{365-x}{x}\right)=0.6368$
On solving, x = 223.92
$\therefore$ Weight of HCl required = 223 mg.