Question

# (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.

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

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 $$\Rightarrow \alpha$$ decreasesas 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

Suggest Corrections

0

Similar questions
View More

People also searched for
View More