How many g moles of HCl will be required to prepare one litre of buffer solution (containing NaCN and HCN) of pH= 8.5 using 0.01 gm formula mass of NaCN. Ka(HCN)=4.1×10–10
A
0.0089 mole
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B
0.0081 mole
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C
0.0082 mole
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D
0.0080 mole
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Solution
The correct option is A 0.0089 mole NaCN+HCl→NaCl+HCN [NaCN]=(0.01−a),[HCN]=a ∴pH=pKa+log[NαCN][HCN] 8.5=log0.01−ααlog4.1×10−10 ∴log0.01−αα=8.5+0.617−10 ∴a=0.0089mole