How many gram-mole of HCl will be required to prepare one litre of buffer solution (containing NaCN and HCl) of pH 8.5 using 0.01 g formula weight of NaCN? K HCN = 4.1 × 10 - 10

Question

How many gram-mole of HCl will be required to prepare one litre of buffer solution (containing NaCN and HCl) of pH 8.5 using 0.01 g formula weight of NaCN? K HCN = 4.1 × 10 - 10  (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

HCN for buffer will be formed by the reaction NaCN + HCl longrightarrow NaCl + HCN mmol of NaCN present initially = ( 0.01/ 49 ) × 1000 = 0.2 Let x mmol of HCl is added so that x mmol of NaCN will be neutralised forming x mmol of HCN. pH = pKa + log ( [ NaCN ]/ [ HCN]) 8.5 = - log ( 4.1 × 10 - 10 ) + log ( 0.2 - x/ x) x = 0.177 mmol

Q. How many gram-mole of HCl will be required to prepare one litre of buffer solution (containing NaCN and HCl) of pH 8.5 using 0.01g formula weight of NaCN? KHCN​=4.1×10−10

HCN for buffer will be formed by the reaction NaCN+HCl⟶NaCl+HCN mmol of NaCN present initially = 490.01​×1000=0.2 Let x mmol of HCl is added so that x mmol of NaCN will be neutralised forming x mmol of HCN. pH=pKa​+log[HCN][NaCN]​ 8.5=−log(4.1×10−10)+logx0.2−x​ x=0.177mmol

Q. How many gram-mole of $H Cl$ will be required to prepare one litre of buffer solution (containing $N a CN$ and $H Cl$ ) of pH $8.5$ using $0.01 g$ formula weight of $N a CN$ ?
$K_{H CN} = 4.1 \times 1 0^{- 10}$