315017
On adding which of the following, the \(\mathrm{pH}\) of \({\text{20}}\) \(\mathrm{mL}\) of \({\text{0}}{\text{.1 N HCl}}\) will not alter?
1 \({\text{1 mL of 1 N HCl}}\)
2 \(20 \mathrm{~mL}\) of distilled water
3 \({\text{1 mL of 0}}{\text{.1 N NaOH}}\)
4 \(500 \mathrm{~mL}\) of \(\mathrm{HCl}\) of \(\mathrm{pH}=1\)
Explanation:
\(\mathrm {0.1 \mathrm{~N} \mathrm{HCl}}\) has \(\mathrm {\mathrm{pH}=1}\) and adding an other \(\mathrm {500 \mathrm{~mL}}\) of \(\mathrm {\mathrm{HCl}}\) of same \(\mathrm {\mathrm{pH}}\) will not alter the \(\mathrm {\mathrm{pH}}\).
CHXI07:EQUILIBRIUM
315018
\(\mathrm{NaOH}\) is a strong base. What will be \(\mathrm{pH}\) of \({\text{5}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 2 }}}}{\text{M }}{\mkern 1mu} {\text{NaOH}}\) solution? \({\text{(log2 = 0}}{\text{.3)}}\)
1 14.00
2 13.70
3 13.00
4 12.70
Explanation:
Given, \(\mathrm {\left[\mathrm{OH}^{-}\right]=5 \times 10^{-2}}\) \(\rm { \therefore p O H=-\log 5 \times 10^{-2} \\ =-\log 5+2 \log 10=1.30}\) \(\mathrm {\because p H+p O H=14}\) \(\mathrm {\because p H=14-p O H}\) \(\mathrm {=14-1.30=12.70}\)
JEE - 2013
CHXI07:EQUILIBRIUM
315019
0.01 mol NaOH is added to 10 litres of water. The pH of the solution is ____ .
315020
If \(\mathrm{pH}\) of \(\mathrm{NaOH}\) solution is 12.0 . The \(\mathrm{pH}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution of same molarity will be
1 2.0
2 12.0
3 1.7
4 10.0387
Explanation:
\({\text{NaOH pH}} = 12,\,\,\therefore {\text{pOH}} = 2\) \(\left[\mathrm{OH}^{-}\right]=10^{-2}\) Molarity of \(\mathrm{NaOH}=10^{-2}\) For \(\mathrm{H}_{2} \mathrm{SO}_{4}\), molarity \(=10^{-2}\) and Normality \(=2 \times 10^{-2}\) \(\therefore\left[\mathrm{H}^{+}\right]=2 \times 10^{-2}\) \(\mathrm{pH}=2-\log 2=1.7\)
315017
On adding which of the following, the \(\mathrm{pH}\) of \({\text{20}}\) \(\mathrm{mL}\) of \({\text{0}}{\text{.1 N HCl}}\) will not alter?
1 \({\text{1 mL of 1 N HCl}}\)
2 \(20 \mathrm{~mL}\) of distilled water
3 \({\text{1 mL of 0}}{\text{.1 N NaOH}}\)
4 \(500 \mathrm{~mL}\) of \(\mathrm{HCl}\) of \(\mathrm{pH}=1\)
Explanation:
\(\mathrm {0.1 \mathrm{~N} \mathrm{HCl}}\) has \(\mathrm {\mathrm{pH}=1}\) and adding an other \(\mathrm {500 \mathrm{~mL}}\) of \(\mathrm {\mathrm{HCl}}\) of same \(\mathrm {\mathrm{pH}}\) will not alter the \(\mathrm {\mathrm{pH}}\).
CHXI07:EQUILIBRIUM
315018
\(\mathrm{NaOH}\) is a strong base. What will be \(\mathrm{pH}\) of \({\text{5}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 2 }}}}{\text{M }}{\mkern 1mu} {\text{NaOH}}\) solution? \({\text{(log2 = 0}}{\text{.3)}}\)
1 14.00
2 13.70
3 13.00
4 12.70
Explanation:
Given, \(\mathrm {\left[\mathrm{OH}^{-}\right]=5 \times 10^{-2}}\) \(\rm { \therefore p O H=-\log 5 \times 10^{-2} \\ =-\log 5+2 \log 10=1.30}\) \(\mathrm {\because p H+p O H=14}\) \(\mathrm {\because p H=14-p O H}\) \(\mathrm {=14-1.30=12.70}\)
JEE - 2013
CHXI07:EQUILIBRIUM
315019
0.01 mol NaOH is added to 10 litres of water. The pH of the solution is ____ .
315020
If \(\mathrm{pH}\) of \(\mathrm{NaOH}\) solution is 12.0 . The \(\mathrm{pH}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution of same molarity will be
1 2.0
2 12.0
3 1.7
4 10.0387
Explanation:
\({\text{NaOH pH}} = 12,\,\,\therefore {\text{pOH}} = 2\) \(\left[\mathrm{OH}^{-}\right]=10^{-2}\) Molarity of \(\mathrm{NaOH}=10^{-2}\) For \(\mathrm{H}_{2} \mathrm{SO}_{4}\), molarity \(=10^{-2}\) and Normality \(=2 \times 10^{-2}\) \(\therefore\left[\mathrm{H}^{+}\right]=2 \times 10^{-2}\) \(\mathrm{pH}=2-\log 2=1.7\)
315017
On adding which of the following, the \(\mathrm{pH}\) of \({\text{20}}\) \(\mathrm{mL}\) of \({\text{0}}{\text{.1 N HCl}}\) will not alter?
1 \({\text{1 mL of 1 N HCl}}\)
2 \(20 \mathrm{~mL}\) of distilled water
3 \({\text{1 mL of 0}}{\text{.1 N NaOH}}\)
4 \(500 \mathrm{~mL}\) of \(\mathrm{HCl}\) of \(\mathrm{pH}=1\)
Explanation:
\(\mathrm {0.1 \mathrm{~N} \mathrm{HCl}}\) has \(\mathrm {\mathrm{pH}=1}\) and adding an other \(\mathrm {500 \mathrm{~mL}}\) of \(\mathrm {\mathrm{HCl}}\) of same \(\mathrm {\mathrm{pH}}\) will not alter the \(\mathrm {\mathrm{pH}}\).
CHXI07:EQUILIBRIUM
315018
\(\mathrm{NaOH}\) is a strong base. What will be \(\mathrm{pH}\) of \({\text{5}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 2 }}}}{\text{M }}{\mkern 1mu} {\text{NaOH}}\) solution? \({\text{(log2 = 0}}{\text{.3)}}\)
1 14.00
2 13.70
3 13.00
4 12.70
Explanation:
Given, \(\mathrm {\left[\mathrm{OH}^{-}\right]=5 \times 10^{-2}}\) \(\rm { \therefore p O H=-\log 5 \times 10^{-2} \\ =-\log 5+2 \log 10=1.30}\) \(\mathrm {\because p H+p O H=14}\) \(\mathrm {\because p H=14-p O H}\) \(\mathrm {=14-1.30=12.70}\)
JEE - 2013
CHXI07:EQUILIBRIUM
315019
0.01 mol NaOH is added to 10 litres of water. The pH of the solution is ____ .
315020
If \(\mathrm{pH}\) of \(\mathrm{NaOH}\) solution is 12.0 . The \(\mathrm{pH}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution of same molarity will be
1 2.0
2 12.0
3 1.7
4 10.0387
Explanation:
\({\text{NaOH pH}} = 12,\,\,\therefore {\text{pOH}} = 2\) \(\left[\mathrm{OH}^{-}\right]=10^{-2}\) Molarity of \(\mathrm{NaOH}=10^{-2}\) For \(\mathrm{H}_{2} \mathrm{SO}_{4}\), molarity \(=10^{-2}\) and Normality \(=2 \times 10^{-2}\) \(\therefore\left[\mathrm{H}^{+}\right]=2 \times 10^{-2}\) \(\mathrm{pH}=2-\log 2=1.7\)
315017
On adding which of the following, the \(\mathrm{pH}\) of \({\text{20}}\) \(\mathrm{mL}\) of \({\text{0}}{\text{.1 N HCl}}\) will not alter?
1 \({\text{1 mL of 1 N HCl}}\)
2 \(20 \mathrm{~mL}\) of distilled water
3 \({\text{1 mL of 0}}{\text{.1 N NaOH}}\)
4 \(500 \mathrm{~mL}\) of \(\mathrm{HCl}\) of \(\mathrm{pH}=1\)
Explanation:
\(\mathrm {0.1 \mathrm{~N} \mathrm{HCl}}\) has \(\mathrm {\mathrm{pH}=1}\) and adding an other \(\mathrm {500 \mathrm{~mL}}\) of \(\mathrm {\mathrm{HCl}}\) of same \(\mathrm {\mathrm{pH}}\) will not alter the \(\mathrm {\mathrm{pH}}\).
CHXI07:EQUILIBRIUM
315018
\(\mathrm{NaOH}\) is a strong base. What will be \(\mathrm{pH}\) of \({\text{5}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 2 }}}}{\text{M }}{\mkern 1mu} {\text{NaOH}}\) solution? \({\text{(log2 = 0}}{\text{.3)}}\)
1 14.00
2 13.70
3 13.00
4 12.70
Explanation:
Given, \(\mathrm {\left[\mathrm{OH}^{-}\right]=5 \times 10^{-2}}\) \(\rm { \therefore p O H=-\log 5 \times 10^{-2} \\ =-\log 5+2 \log 10=1.30}\) \(\mathrm {\because p H+p O H=14}\) \(\mathrm {\because p H=14-p O H}\) \(\mathrm {=14-1.30=12.70}\)
JEE - 2013
CHXI07:EQUILIBRIUM
315019
0.01 mol NaOH is added to 10 litres of water. The pH of the solution is ____ .
315020
If \(\mathrm{pH}\) of \(\mathrm{NaOH}\) solution is 12.0 . The \(\mathrm{pH}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) solution of same molarity will be
1 2.0
2 12.0
3 1.7
4 10.0387
Explanation:
\({\text{NaOH pH}} = 12,\,\,\therefore {\text{pOH}} = 2\) \(\left[\mathrm{OH}^{-}\right]=10^{-2}\) Molarity of \(\mathrm{NaOH}=10^{-2}\) For \(\mathrm{H}_{2} \mathrm{SO}_{4}\), molarity \(=10^{-2}\) and Normality \(=2 \times 10^{-2}\) \(\therefore\left[\mathrm{H}^{+}\right]=2 \times 10^{-2}\) \(\mathrm{pH}=2-\log 2=1.7\)
