Question

Calculate the weight (in $mg$) of $HCl$ that should be added to $100ml$ of $0.1MBOH$ to make a buffer solution of $pOH=6$.
$K_b\left(BOH\right)=1\times {10}^{-8}$ (assume there is no change in volume on addition of $HCl$)

• A. 3.6
• B. 4.6
• C. 5.6
• D. 6.6

Answer 1

The correct option is A 3.6

Let the mass of $HCl$ added be $xmg$
$mmol$ of $HCl$ is $\frac{\text{Mass (in mg)}}{\text{Molar mass}}=\frac{x}{36.5}$
$mmol$ of $BOH$ is $100\times 0.1=10$
The $mmol$ of each species is calculated below :
$BOH+HCl\to BCl+H_{2}O$
$t=0:(100\times 0.1=10)\frac{x}{36.5}t=t_{eq}:10-\frac{x}{36.5}0\frac{x}{36.5}=\frac{365-x}{36.5}$

This constitutes basic buffer. For basic buffer, the $pOH$ given by:

$pOH=p{K}_b+\log \frac{\left[BCl\right]}{\left[BOH\right]}pOH=-\log [1\times {10}^{-8}]+\log \left(\frac{x\times 36.5}{36.5\times (365-x)}\right)6=8+\log \left(\frac{x}{365-x}\right)-2=\log \left(\frac{x}{365-x}\right){10}^{-2}=\left(\frac{x}{365-x}\right)3.65-0.01x=x1.01x=3.65x=3.61\approx 3.6$

$\therefore$ Weight of $HCl$ required = 3.6 mg.