315005
Statement A : The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid. Statement B : Hydrogen sulphide is a weak acid.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\({{\text{H}}_{\text{2}}}{\text{S}}\) is a weak acid and also a weak electrolyte and its ionisation is suppressed, when small amount of strong electrolyte like \(\mathrm {\mathrm{HCl}}\) is added due to common ion effect. \(\rm {\mathrm{H}_{2} \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}}\) \(\rm {\mathrm{HCl} \rightarrow \underset{\substack{\text { Common } \\ \text { ion }}}{\mathrm{H}^{+}}+\mathrm{Cl}^{-}}\) So, the option (3) is correct.
CHXI07:EQUILIBRIUM
315006
Calculate the \(\mathrm{pH}\) of a solution which contains \({\text{10}}\,\,{\text{ml}}\) of \({\text{1 M}}{\mkern 1mu} \,{\text{HCl}}\) and \({\text{10}}\,\,{\text{ml}}\) of \({\text{2 M}}{\mkern 1mu} \,{\text{NaOH}}\)
315008
The \(\mathrm{pH}\) of a solution prepared by mixing \(2.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) solution of \(\mathrm{pH} 3.0\) and \(3.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\) of \(\mathrm{pH} 10.0\) is
1 2.5
2 3.5
3 5.5
4 6.5
Explanation:
\(\because \mathrm{pH}\) of \(\mathrm{HCl}\) solution \(=3.0\) \(\therefore\left[\mathrm{H}^{+}\right]\)in \(\mathrm{HCl}\) solution \(=1 \times 10^{-3}\) \(\because \mathrm{pH}\) of \(\mathrm{NaOH}\) solution \(=10.0\) \(\therefore\left[\mathrm{H}^{+}\right]\)in \(\mathrm{NaOH}\) solution \(=1 \times 10^{-10}\) \(\left[\mathrm{OH}^{-}\right]\)in \(\mathrm{NaOH}\) solution\(=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=10^{-4}\) Milliequivalents of \(\mathrm{HCl}\) \(=\mathrm{N}_{1} \mathrm{~V}_{1}=2.0 \times 1 \times 10^{-3}=2.0 \times 10^{-3}\) Milliequivalents of \(\mathrm{NaOH}\) \(=3.0 \times 1 \times 10^{-4}=3.0 \times 10^{-4}\) Since, milliequivalents of \(\mathrm{HCl}\) are in excess, the milliequivalents of \(\left[\mathrm{H}^{+}\right]\)in mixture \(\begin{aligned}& =\left(2.0 \times 10^{-3}\right)-\left(3.0 \times 10^{-4}\right) \\& =1.7 \times 10^{-3}\end{aligned}\) Concentration of in \(\left[\mathrm{H}^{+}\right]\)i mixture \(\begin{aligned}&=\dfrac{1.7 \times 10^{-3}}{3+2}=3.4 \times 10^{-4} \\& \mathrm{pH} \text { of mixture }=-\log \left[\mathrm{H}^{+}\right] \\&=-\log \left(3.4 \times 10^{-4}\right) \\&=3.46 \sim 3.5\end{aligned}\)
315005
Statement A : The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid. Statement B : Hydrogen sulphide is a weak acid.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\({{\text{H}}_{\text{2}}}{\text{S}}\) is a weak acid and also a weak electrolyte and its ionisation is suppressed, when small amount of strong electrolyte like \(\mathrm {\mathrm{HCl}}\) is added due to common ion effect. \(\rm {\mathrm{H}_{2} \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}}\) \(\rm {\mathrm{HCl} \rightarrow \underset{\substack{\text { Common } \\ \text { ion }}}{\mathrm{H}^{+}}+\mathrm{Cl}^{-}}\) So, the option (3) is correct.
CHXI07:EQUILIBRIUM
315006
Calculate the \(\mathrm{pH}\) of a solution which contains \({\text{10}}\,\,{\text{ml}}\) of \({\text{1 M}}{\mkern 1mu} \,{\text{HCl}}\) and \({\text{10}}\,\,{\text{ml}}\) of \({\text{2 M}}{\mkern 1mu} \,{\text{NaOH}}\)
315008
The \(\mathrm{pH}\) of a solution prepared by mixing \(2.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) solution of \(\mathrm{pH} 3.0\) and \(3.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\) of \(\mathrm{pH} 10.0\) is
1 2.5
2 3.5
3 5.5
4 6.5
Explanation:
\(\because \mathrm{pH}\) of \(\mathrm{HCl}\) solution \(=3.0\) \(\therefore\left[\mathrm{H}^{+}\right]\)in \(\mathrm{HCl}\) solution \(=1 \times 10^{-3}\) \(\because \mathrm{pH}\) of \(\mathrm{NaOH}\) solution \(=10.0\) \(\therefore\left[\mathrm{H}^{+}\right]\)in \(\mathrm{NaOH}\) solution \(=1 \times 10^{-10}\) \(\left[\mathrm{OH}^{-}\right]\)in \(\mathrm{NaOH}\) solution\(=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=10^{-4}\) Milliequivalents of \(\mathrm{HCl}\) \(=\mathrm{N}_{1} \mathrm{~V}_{1}=2.0 \times 1 \times 10^{-3}=2.0 \times 10^{-3}\) Milliequivalents of \(\mathrm{NaOH}\) \(=3.0 \times 1 \times 10^{-4}=3.0 \times 10^{-4}\) Since, milliequivalents of \(\mathrm{HCl}\) are in excess, the milliequivalents of \(\left[\mathrm{H}^{+}\right]\)in mixture \(\begin{aligned}& =\left(2.0 \times 10^{-3}\right)-\left(3.0 \times 10^{-4}\right) \\& =1.7 \times 10^{-3}\end{aligned}\) Concentration of in \(\left[\mathrm{H}^{+}\right]\)i mixture \(\begin{aligned}&=\dfrac{1.7 \times 10^{-3}}{3+2}=3.4 \times 10^{-4} \\& \mathrm{pH} \text { of mixture }=-\log \left[\mathrm{H}^{+}\right] \\&=-\log \left(3.4 \times 10^{-4}\right) \\&=3.46 \sim 3.5\end{aligned}\)
315005
Statement A : The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid. Statement B : Hydrogen sulphide is a weak acid.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\({{\text{H}}_{\text{2}}}{\text{S}}\) is a weak acid and also a weak electrolyte and its ionisation is suppressed, when small amount of strong electrolyte like \(\mathrm {\mathrm{HCl}}\) is added due to common ion effect. \(\rm {\mathrm{H}_{2} \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}}\) \(\rm {\mathrm{HCl} \rightarrow \underset{\substack{\text { Common } \\ \text { ion }}}{\mathrm{H}^{+}}+\mathrm{Cl}^{-}}\) So, the option (3) is correct.
CHXI07:EQUILIBRIUM
315006
Calculate the \(\mathrm{pH}\) of a solution which contains \({\text{10}}\,\,{\text{ml}}\) of \({\text{1 M}}{\mkern 1mu} \,{\text{HCl}}\) and \({\text{10}}\,\,{\text{ml}}\) of \({\text{2 M}}{\mkern 1mu} \,{\text{NaOH}}\)
315008
The \(\mathrm{pH}\) of a solution prepared by mixing \(2.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) solution of \(\mathrm{pH} 3.0\) and \(3.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\) of \(\mathrm{pH} 10.0\) is
1 2.5
2 3.5
3 5.5
4 6.5
Explanation:
\(\because \mathrm{pH}\) of \(\mathrm{HCl}\) solution \(=3.0\) \(\therefore\left[\mathrm{H}^{+}\right]\)in \(\mathrm{HCl}\) solution \(=1 \times 10^{-3}\) \(\because \mathrm{pH}\) of \(\mathrm{NaOH}\) solution \(=10.0\) \(\therefore\left[\mathrm{H}^{+}\right]\)in \(\mathrm{NaOH}\) solution \(=1 \times 10^{-10}\) \(\left[\mathrm{OH}^{-}\right]\)in \(\mathrm{NaOH}\) solution\(=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=10^{-4}\) Milliequivalents of \(\mathrm{HCl}\) \(=\mathrm{N}_{1} \mathrm{~V}_{1}=2.0 \times 1 \times 10^{-3}=2.0 \times 10^{-3}\) Milliequivalents of \(\mathrm{NaOH}\) \(=3.0 \times 1 \times 10^{-4}=3.0 \times 10^{-4}\) Since, milliequivalents of \(\mathrm{HCl}\) are in excess, the milliequivalents of \(\left[\mathrm{H}^{+}\right]\)in mixture \(\begin{aligned}& =\left(2.0 \times 10^{-3}\right)-\left(3.0 \times 10^{-4}\right) \\& =1.7 \times 10^{-3}\end{aligned}\) Concentration of in \(\left[\mathrm{H}^{+}\right]\)i mixture \(\begin{aligned}&=\dfrac{1.7 \times 10^{-3}}{3+2}=3.4 \times 10^{-4} \\& \mathrm{pH} \text { of mixture }=-\log \left[\mathrm{H}^{+}\right] \\&=-\log \left(3.4 \times 10^{-4}\right) \\&=3.46 \sim 3.5\end{aligned}\)
315005
Statement A : The ionisation of hydrogen sulphide in water is low in the presence of hydrochloric acid. Statement B : Hydrogen sulphide is a weak acid.
1 Statement A is correct but Statement B is incorrect.
2 Statement A is incorrect but Statement B is correct.
3 Both statements are correct.
4 Both Statements are incorrect.
Explanation:
\({{\text{H}}_{\text{2}}}{\text{S}}\) is a weak acid and also a weak electrolyte and its ionisation is suppressed, when small amount of strong electrolyte like \(\mathrm {\mathrm{HCl}}\) is added due to common ion effect. \(\rm {\mathrm{H}_{2} \mathrm{~S} \rightleftharpoons 2 \mathrm{H}^{+}+\mathrm{S}^{2-}}\) \(\rm {\mathrm{HCl} \rightarrow \underset{\substack{\text { Common } \\ \text { ion }}}{\mathrm{H}^{+}}+\mathrm{Cl}^{-}}\) So, the option (3) is correct.
CHXI07:EQUILIBRIUM
315006
Calculate the \(\mathrm{pH}\) of a solution which contains \({\text{10}}\,\,{\text{ml}}\) of \({\text{1 M}}{\mkern 1mu} \,{\text{HCl}}\) and \({\text{10}}\,\,{\text{ml}}\) of \({\text{2 M}}{\mkern 1mu} \,{\text{NaOH}}\)
315008
The \(\mathrm{pH}\) of a solution prepared by mixing \(2.0 \mathrm{~mL}\) of \(\mathrm{HCl}\) solution of \(\mathrm{pH} 3.0\) and \(3.0 \mathrm{~mL}\) of \(\mathrm{NaOH}\) of \(\mathrm{pH} 10.0\) is
1 2.5
2 3.5
3 5.5
4 6.5
Explanation:
\(\because \mathrm{pH}\) of \(\mathrm{HCl}\) solution \(=3.0\) \(\therefore\left[\mathrm{H}^{+}\right]\)in \(\mathrm{HCl}\) solution \(=1 \times 10^{-3}\) \(\because \mathrm{pH}\) of \(\mathrm{NaOH}\) solution \(=10.0\) \(\therefore\left[\mathrm{H}^{+}\right]\)in \(\mathrm{NaOH}\) solution \(=1 \times 10^{-10}\) \(\left[\mathrm{OH}^{-}\right]\)in \(\mathrm{NaOH}\) solution\(=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=10^{-4}\) Milliequivalents of \(\mathrm{HCl}\) \(=\mathrm{N}_{1} \mathrm{~V}_{1}=2.0 \times 1 \times 10^{-3}=2.0 \times 10^{-3}\) Milliequivalents of \(\mathrm{NaOH}\) \(=3.0 \times 1 \times 10^{-4}=3.0 \times 10^{-4}\) Since, milliequivalents of \(\mathrm{HCl}\) are in excess, the milliequivalents of \(\left[\mathrm{H}^{+}\right]\)in mixture \(\begin{aligned}& =\left(2.0 \times 10^{-3}\right)-\left(3.0 \times 10^{-4}\right) \\& =1.7 \times 10^{-3}\end{aligned}\) Concentration of in \(\left[\mathrm{H}^{+}\right]\)i mixture \(\begin{aligned}&=\dfrac{1.7 \times 10^{-3}}{3+2}=3.4 \times 10^{-4} \\& \mathrm{pH} \text { of mixture }=-\log \left[\mathrm{H}^{+}\right] \\&=-\log \left(3.4 \times 10^{-4}\right) \\&=3.46 \sim 3.5\end{aligned}\)
