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(a) When 100 ml of 0.1 M NaCN solution is titrated with 0.1 M HCI solution the variation of pH of solution with volume of HCI added will be : 
(b) Variation of degree of dissociation $$ \alpha $$ with concentration for a weak electrolyte at a particular temperature is best represnted by : 
(c)  M acetic acid solution is titrated against  M NaOH solution. the difference in pH between 1/4 and 3/4 stages of neutralization of acid will be 2 log 3.
1745217_0a1fd8bbe6e54c1dacc1517bac441531.PNG


A
T,F,T
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B
F,F,F
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C
T,T,T
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D
F,T,F
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Solution

The correct option is A $$ T , F, T $$
(a) Initially pH will decrease fast, then slowly due to buffer formation and then will decrease fast as buffer action diminishes.

(b) for a weak electrolyte
$$ K_a = \frac {C \alpha^2}{( 1 - \alpha) } $$ when $$ \alpha << 1 then, \alpha = \sqrt { \frac {k_a}{C} } $$
as C increases $$ \Rightarrow \alpha $$ decreases
as C is tending to zero $$ \Rightarrow  \alpha $$ will be unity 

(c) At 1/4th neutralisation
$$ CH_3COH +NaOH  \rightarrow CH_3COONa +H_2O $$
$$ ( 0.1 \times \frac {3}{4} ) \quad \quad \quad \quad \quad \,\,\,( 0.1 \times \frac {1}{4} )$$
$$ pH = pK_a + log  \frac {[CH_3COO^-]}{[CH_3COOH[} = pK_a +log ( \frac {1}{3} ) $$

At 3/4th neutralisation
$$pH = pK_a + log 3 $$
So difference in pH = $$ \Delta (pH) = log 3 - log \frac {1}{3} = 2 log 3 $$

Chemistry

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