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Standard XII

Chemistry

Acidity of Alcohols and Phenols

A handbook st...

Question

A handbook states that the solubility of methylamine ${C H}_{3} {N H}_{2} (g)$ in water at $1$ atm pressure at $25^{o} C$ is $959$ volumes of ${C H}_{3} {N H}_{2} (g)$ per volume of water $(p k_{b} = 3.39)$ .

If the molarity of $N a O H (a q .)$ required to yield the same $p H$ is $125 \times 10^{- x}$ , then what is the value of x?

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Solution

$P V = n R T$
$1 \times 0.959 = n \times 0.0821 \times 298$
$n = 0.03919$
Volume of $H_{2} O = 1 m L$ (per volume of $H_{2} O$ )
$C = \frac{n}{V} = \frac{0.03919}{10^{- 3}} = 39.19 M$
$p k_{b} = 3.39$ $\Rightarrow$ $k_{b} = 4 \times 10^{- 4}$
$[{O H}^{-}] = \sqrt{K_{b} C} = 0.1252 M$
$p O H = 0.9023$
$p H = 13.097$
$M = 0.1252$ or $125 \times 10^{- 3}$ for $N a O H$ will produce same pH.
Hence, the value of $x$ is $3$ .

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A handbook states that the solubility of methylamine CH3NH2(g) in water at 1 atm pressure at 25oC is 959 volumes of CH3NH2(g) per volume of water (pkb=3.39).If the molarity of NaOH(aq.) required to yield the same pH is 125×10−x, then what is the value of x?

PV=nRT1×0.959=n×0.0821×298n=0.03919Volume of H2O=1 mL (per volume of H2O)C=nV=0.0391910−3=39.19Mpkb=3.39 ⇒ kb=4×10−4[OH−]=√KbC=0.1252MpOH=0.9023 pH=13.097M=0.1252 or 125×10−3 for NaOH will produce same pH.Hence, the value of x is 3.