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Auto Protolysis of Water

A 0.004 M s...

Question

A $0.004$ M solution of $N a_{2} S O_{4}$ is isotonic with a $0.010$ M solution of glucose at same temperature. The apparent degree of dissociation of $N a_{2} S O_{4}$ is:

25

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50

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75

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85

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Solution

The correct option is C $75$ %
The reaction for the dissociation of isotonic solutions is as given below.
$N a_{2} S O_{4} 0.004 - x ⇌ 2 N a^{+} 2 x + S O_{4}^{2 -} x$
Both the solutions are isotonic.
Hence, $(0.004 - x) + 2 x - x = 0.01$ or $x = 3 \times 10^{- 3}$ .
Percentage dissociation $= \frac{3 \times 10^{- 3}}{0.004} \times 100 = 75$ %.

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A 0.004 M solution of Na2SO4 is isotonic with a 0.010 M solution of glucose at same temperature. The apparent degree of dissociation of Na2SO4 is:

The correct option is C 75%The reaction for the dissociation of isotonic solutions is as given below.Na2SO40.004−x⇌2Na+2x+SO2−4xBoth the solutions are isotonic.Hence, (0.004−x)+2x−x=0.01 or x=3×10−3. Percentage dissociation =3×10−30.004×100=75%.

A $0.004$ M solution of $N a_{2} S O_{4}$ is isotonic with a $0.010$ M solution of glucose at same temperature. The apparent degree of dissociation of $N a_{2} S O_{4}$ is: