Question

How many mole of $HCN$ will be required to prepare one litre buffer solution of $pH =10.4$ using $0.01$ mol $NaCN$ ? (Given $\left.Ka ( HCN )=4 \times 10^{-10}\right)$

• A. $2 \times 10^{-2} mol$
• B. $1 \times 10^{-3} mol$
• C. $4 \times 10^{-4} mol$
• D. $9 \times 10^{-3} mol$

Answer 1

Answer: (B)

$pH = pk _{ a }+\log \left(\frac{ Salt }{ Acid }\right)$
$10.4=9.4+\log \left(\frac{0.01}{ x }\right)$
$1=\log \frac{0.01}{ x } ; 10=\frac{0.01}{ x }$
$x =0.001=10^{-3}$