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  • Consider the following statements: (a) The pH of a mixture containing 400mL of 0.1 M H2SO4 and 400ml of 0.1M NaOH will be approximately 1.3. (b) Ionic product of water is temperature dependent. (c) A monobasic acid with 

Consider the following statements: (a) The pH of a mixture containing 400mL of 0.1 M H2SO4 and 400ml of 0.1M NaOH will be approximately 1.3. (b) Ionic product of water is temperature dependent. (c) A monobasic acid with K_{a}=10^{-5} has a pH=5. The degree of dissociation of this acid is 50%. (d) The Le Chatelier's principle is not applicable to common-ion effect. The correct statements are :


Option: 1 (a), (b) and (d)
Option: 2 (a), (b) and (c)
Option: 3 (b) and (c)
Option: 4 (a) and (b)
 

Answers (1)

best_answer

(a) H_2SO_4+2NaOH\rightarrow Na_2SO_4+2H_2O

Mole of H2SO4 = 400 x 0.1 = 40     

Mole of NaOH = 400 x 0.1 = 40      

for 40 mol of H2SO4 , Need 80 mol of NaOH. But NaOH is 40 mol only.

So, NaOH is a limiting reagent.

Hence, 40 moles of NaOH will react with only 20 moles of H2SO4. 

\mathrm{[H^{+}]=\frac{20\times 2}{800(\textup{Total volume})}=\frac{1}{20}}

\mathrm{pH=-\log \left (\frac{1}{20} \right )=1.3}

(b) \mathrm{\log \frac{K_{w2}}{K_{w1}}=\frac{\Delta H}{2.303R}\left [\frac{1}{T_1}-\frac{1}{T_2} \right ]}

Thus, the ionic product of water is temperature-dependent.

 

(c) \mathrm{K_a=10^{-5}\textup{ and }pH=5\Rightarrow [H^{+}]=10^{-5}}

Now, 

\\\mathrm{K_a=\frac{C\alpha ^{2}}{(1-\alpha )}}\\\\\\ \mathrm{K_a=\frac{[H^{+}]\alpha }{(1-\alpha )}}

10^{-5}=\frac{10^{-5}\alpha }{(1-\alpha )}

\mathrm{\alpha =50\%\textup{ or 1/2}}

 

(d) Le-chatelier's principle is applicable to the common ion effect. 

So, statement (d) is wrong. 

Therefore, option(2) is correct.

Posted by

Ritika Jonwal

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