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Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium

If freezing point of 0.1 M aqueous weak monobasic acid is $-0.2046^\circ C$, then $pH$ of solution is [$K_f (H_2O) = 1.86 kg mol^{-1}$]

(a) 1

(b) 2

(c) 1.3

(d) 1.7

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1 Answer

Answer: 2
 
$\Delta T_f = i K_f m$
$i = \frac{\Delta T_f}{K_f m} = \frac{0.2046}{1.86 \times 0.1} = \frac{204.6}{186}$ ............(i)
 
$HA \rightleftharpoons H^+ + A^-$
Initially, [HA] = 0.1 M
At Equilibrium, [HA] = 0.1 -x, [$H^+$] = x, [$A^-$] = x
$\therefore i = \frac{0.1 - x + 2x}{0.1} = \frac{0.1 + x}{0.1} = 1 + 10x$...............(ii)
 
From equation (i) and (i), $x = 10^{-2}$
 
$\therefore pH = - log [H^+] = -log 10^{-2} = 2$

 

answered Nov 29, 2013 by mosymeow_1
 
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