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Standard XII

Chemistry

Basic Buffer Action

How many gram...

Question

How many gram moles of $H C l$ will be required to prepare one litre of a buffer solution (containing $N a C N$ and $H C N$ ) of pH $8.5$ using $0.10 g$ formula mass of $N a C N$ ? $K_{a}$ for $H C N = 4.1 \times 10^{- 10}$ .

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Solution

Let $a$ mole of $H C l$ be added. It will combine with $N a C N$ to form $H C N$
$N a C N + H C l ⟶ N a C l + H C N$
$a$ $a$ $a$
$[N a C N] = (0.01 - a); [H C N] = a$
Applying the equation
$p H = log \frac{[N a C N]}{[H C N]} - log K_{a}$
$8.5 = log \frac{0.01 - a}{a} - log 4.1 \times 10^{- 10}$
So, $a = \frac{0.01}{1.1296} = 0.0089 m o l e$

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Q. 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.

K_{a} (H C N) = 4.1 \times 10^{- 10}

How many gram moles of $H C l$ will be required to prepare one litre of a buffer solution (containing $N a C N$ and $H C N$ ) of $p H 8.5$ using $0.01 g r a m$ formula weight of $N a C N$ ?

$K_{a} (H C N)$ = $4.1 \times 10^{- 10}$

Q. How many mole of

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How many gram moles of HCl will be required to prepare one litre of a buffer solution (containing NaCN and HCN) of pH 8.5 using 0.10g formula mass of NaCN? Ka for HCN=4.1×10−10.

Let a mole of HCl be added. It will combine with NaCN to form HCNNaCN+HCl⟶NaCl+HCNa a a[NaCN]=(0.01−a);[HCN]=aApplying the equationpH=log[NaCN][HCN]−logKa8.5=log0.01−aa−log4.1×10−10So, a=0.011.1296=0.0089mole

How many gram moles of $H C l$ will be required to prepare one litre of a buffer solution (containing $N a C N$ and $H C N$ ) of pH $8.5$ using $0.10 g$ formula mass of $N a C N$ ? $K_{a}$ for $H C N = 4.1 \times 10^{- 10}$ .