Question #a4e04

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Question #a4e04

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Answer 1 · 2017-05-26T05:19:12

The $p K_{a}$ $pK_a$ of $H N O_{2} = 3.25$ $HNO_2=3.25$

Let $x$ $x$ L of 0.100M solution of. $H N O_{2}$ $HNO_2$ be mixed with $(1 - x)$ $(1-x)$ L of. $N a N O_{2}$ $NaNO_2$ to make $1$ $1$ L of buffer solution.

Now $x$ $x$ L of 0.100M solution of $H N O_{2}$ $HNO_2$ contains $0.1 x$ $0.1x$ mol $H N O_{2}$ $HNO_2$ in 1 L mixture

and $(1 - x)$ $(1-x)$ L of. $N a N O_{2}$ $NaNO_2$ solution contains $0.2 (1 - x)$ $0.2(1-x)$ mol $N a N O_{2}$ $NaNO_2$ in 1 L mixture.

Now Using the Henderson-Hasselbalch equation

$p H = p K_{a} + {log}_{10} (\frac{[N a N O_{2}]}{[H N O_{2}]})$ $pH=pK_a+log_10(([NaNO_2])/([HNO_2]))$

$\Rightarrow 3.35 = 3.25 + {log}_{10} (\frac{0.2 (1 - x)}{0.1 x})$ $=>3.35=3.25+log_10((0.2(1-x))/(0.1x))$

$\Rightarrow {log}_{10} (\frac{0.2 (1 - x)}{0.1 x}) = 0.01$ $=>log_10((0.2(1-x))/(0.1x))=0.01$

$\Rightarrow \frac{2 (1 - x)}{x} = 10^{0.01} \approx 1.02$ $=>(2(1-x))/x=10^(0.01)~~1.02$

$\Rightarrow \frac{1}{x} - \frac{x}{x} = 0.51$ $=>1/x-x/x=0.51$

$\Rightarrow \frac{1}{x} - 1 = 0.51$ $=>1/x-1=0.51$

$\Rightarrow \frac{1}{x} = 1.51$ $=>1/x=1.51$

$\Rightarrow x = \frac{1}{1.51} \approx \frac{2}{3}$ $=>x=1/1.51~~2/3$

So to prepare $1$ $1$ L buffer $\frac{2}{3}$ $2/3$ L of 0.100M solution of $H N O_{2}$ $HNO_2$ should be mixed with $\frac{1}{3}$ $1/3$ L of 0.200M solution of $N a N O_{2}$ $NaNO_2$ .

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Question #a4e04

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