315056
The number of hydroxyl ions in \(100 \mathrm{~mL}\) of a solution having \({\text{pH = }}10\) is
1 \(1 \times 10^{4}\)
2 \(3.012 \times 10^{4}\)
3 \(6.02 \times 10^{18}\)
4 \(6.023 \times 10^{19}\)
Explanation:
Given, \(\mathrm{pH}=10\) \(\therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10} \mathrm{~mol} / \mathrm{L}\) \(\therefore\left[\mathrm{OH}^{-}\right]=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(\because\) The number of hydroxyl ions present in 1 mole \(=6.023 \times 10^{23}\) \(\therefore\) The number of hydroxyl ion in \(1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(=6.023 \times 10^{23} \times 10^{-4} / \mathrm{L}\) \(=6.023 \times 10^{19} \mathrm{~mol} / \mathrm{L}\) \(\because\) In \(1000 \mathrm{~mL}\), number of hydroxyl ion \(=6.023 \times 10^{19}\) \(\therefore\) In \(100 \mathrm{~mL}\), number of hydroxyl ions \(=\dfrac{6.023 \times 10^{19} \times 100}{1000}=6.02 \times 10^{18}\)
AIIMS - 2009
CHXI07:EQUILIBRIUM
315057
What is the \({\text{pH}}{\mkern 1mu} \) of a \({\text{1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) \({\text{O}}{{\text{H}}^{\text{ - }}}\) solution at \(330 \mathrm{~K}\), if \({{\text{K}}_{\text{w}}}\) at \(330 \mathrm{~K}\) is \(10^{-13.6}\) ?
315058
How many litres of water must be added to 1 litre on an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ?
1 0.9 L
2 2.0 L
3 9.0 L
4 0.1 L
Explanation:
\((\mathrm{pH})_{1}=1 \Rightarrow\left[\mathrm{H}^{+}\right]_{1}=10^{-1}\) \((\mathrm{pH})_{2}=2 \Rightarrow\left[\mathrm{H}^{+}\right]_{2}=10^{-2}\) Dilution law, \({{\rm{M}}_1}\;{{\rm{V}}_{\rm{1}}} = {{\rm{M}}_{\rm{2}}}\;{{\rm{V}}_{\rm{2}}}\) \({10^{ - 1}} \times 1 = {10^{ - 2}} \times {{\rm{V}}_{\rm{2}}}\;\) \({{\rm{V}}_2} = 10\;{\rm{L}}\) Volume of water to be added \(=\mathrm{V}_{2}-\mathrm{V}_{1}=10^{-1}=9 \mathrm{~L}\) (on for every ten times dilution pH of strong acid increases by 1 unit).
CHXI07:EQUILIBRIUM
315059
What is the \(\mathrm{pH}\) of \(0.005\,\,{\text{M}}\,\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\) solution?
1 5.0
2 2.3
3 3.3
4 2.0
Explanation:
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) is dibasic acid, so \(0.005 \mathrm{M}=0.01 \mathrm{~N}\) \(\begin{gathered}=10^{-2} \mathrm{~N} \\\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\=-\log _{10}\left[10^{-2}\right]=2.0\end{gathered}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
315060
Calculate the \({\text{pH of 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{M Ca(OH}}{{\text{)}}_{\text{2}}}\). Given \(\log 2=0.3\)
315056
The number of hydroxyl ions in \(100 \mathrm{~mL}\) of a solution having \({\text{pH = }}10\) is
1 \(1 \times 10^{4}\)
2 \(3.012 \times 10^{4}\)
3 \(6.02 \times 10^{18}\)
4 \(6.023 \times 10^{19}\)
Explanation:
Given, \(\mathrm{pH}=10\) \(\therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10} \mathrm{~mol} / \mathrm{L}\) \(\therefore\left[\mathrm{OH}^{-}\right]=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(\because\) The number of hydroxyl ions present in 1 mole \(=6.023 \times 10^{23}\) \(\therefore\) The number of hydroxyl ion in \(1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(=6.023 \times 10^{23} \times 10^{-4} / \mathrm{L}\) \(=6.023 \times 10^{19} \mathrm{~mol} / \mathrm{L}\) \(\because\) In \(1000 \mathrm{~mL}\), number of hydroxyl ion \(=6.023 \times 10^{19}\) \(\therefore\) In \(100 \mathrm{~mL}\), number of hydroxyl ions \(=\dfrac{6.023 \times 10^{19} \times 100}{1000}=6.02 \times 10^{18}\)
AIIMS - 2009
CHXI07:EQUILIBRIUM
315057
What is the \({\text{pH}}{\mkern 1mu} \) of a \({\text{1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) \({\text{O}}{{\text{H}}^{\text{ - }}}\) solution at \(330 \mathrm{~K}\), if \({{\text{K}}_{\text{w}}}\) at \(330 \mathrm{~K}\) is \(10^{-13.6}\) ?
315058
How many litres of water must be added to 1 litre on an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ?
1 0.9 L
2 2.0 L
3 9.0 L
4 0.1 L
Explanation:
\((\mathrm{pH})_{1}=1 \Rightarrow\left[\mathrm{H}^{+}\right]_{1}=10^{-1}\) \((\mathrm{pH})_{2}=2 \Rightarrow\left[\mathrm{H}^{+}\right]_{2}=10^{-2}\) Dilution law, \({{\rm{M}}_1}\;{{\rm{V}}_{\rm{1}}} = {{\rm{M}}_{\rm{2}}}\;{{\rm{V}}_{\rm{2}}}\) \({10^{ - 1}} \times 1 = {10^{ - 2}} \times {{\rm{V}}_{\rm{2}}}\;\) \({{\rm{V}}_2} = 10\;{\rm{L}}\) Volume of water to be added \(=\mathrm{V}_{2}-\mathrm{V}_{1}=10^{-1}=9 \mathrm{~L}\) (on for every ten times dilution pH of strong acid increases by 1 unit).
CHXI07:EQUILIBRIUM
315059
What is the \(\mathrm{pH}\) of \(0.005\,\,{\text{M}}\,\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\) solution?
1 5.0
2 2.3
3 3.3
4 2.0
Explanation:
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) is dibasic acid, so \(0.005 \mathrm{M}=0.01 \mathrm{~N}\) \(\begin{gathered}=10^{-2} \mathrm{~N} \\\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\=-\log _{10}\left[10^{-2}\right]=2.0\end{gathered}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
315060
Calculate the \({\text{pH of 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{M Ca(OH}}{{\text{)}}_{\text{2}}}\). Given \(\log 2=0.3\)
315056
The number of hydroxyl ions in \(100 \mathrm{~mL}\) of a solution having \({\text{pH = }}10\) is
1 \(1 \times 10^{4}\)
2 \(3.012 \times 10^{4}\)
3 \(6.02 \times 10^{18}\)
4 \(6.023 \times 10^{19}\)
Explanation:
Given, \(\mathrm{pH}=10\) \(\therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10} \mathrm{~mol} / \mathrm{L}\) \(\therefore\left[\mathrm{OH}^{-}\right]=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(\because\) The number of hydroxyl ions present in 1 mole \(=6.023 \times 10^{23}\) \(\therefore\) The number of hydroxyl ion in \(1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(=6.023 \times 10^{23} \times 10^{-4} / \mathrm{L}\) \(=6.023 \times 10^{19} \mathrm{~mol} / \mathrm{L}\) \(\because\) In \(1000 \mathrm{~mL}\), number of hydroxyl ion \(=6.023 \times 10^{19}\) \(\therefore\) In \(100 \mathrm{~mL}\), number of hydroxyl ions \(=\dfrac{6.023 \times 10^{19} \times 100}{1000}=6.02 \times 10^{18}\)
AIIMS - 2009
CHXI07:EQUILIBRIUM
315057
What is the \({\text{pH}}{\mkern 1mu} \) of a \({\text{1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) \({\text{O}}{{\text{H}}^{\text{ - }}}\) solution at \(330 \mathrm{~K}\), if \({{\text{K}}_{\text{w}}}\) at \(330 \mathrm{~K}\) is \(10^{-13.6}\) ?
315058
How many litres of water must be added to 1 litre on an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ?
1 0.9 L
2 2.0 L
3 9.0 L
4 0.1 L
Explanation:
\((\mathrm{pH})_{1}=1 \Rightarrow\left[\mathrm{H}^{+}\right]_{1}=10^{-1}\) \((\mathrm{pH})_{2}=2 \Rightarrow\left[\mathrm{H}^{+}\right]_{2}=10^{-2}\) Dilution law, \({{\rm{M}}_1}\;{{\rm{V}}_{\rm{1}}} = {{\rm{M}}_{\rm{2}}}\;{{\rm{V}}_{\rm{2}}}\) \({10^{ - 1}} \times 1 = {10^{ - 2}} \times {{\rm{V}}_{\rm{2}}}\;\) \({{\rm{V}}_2} = 10\;{\rm{L}}\) Volume of water to be added \(=\mathrm{V}_{2}-\mathrm{V}_{1}=10^{-1}=9 \mathrm{~L}\) (on for every ten times dilution pH of strong acid increases by 1 unit).
CHXI07:EQUILIBRIUM
315059
What is the \(\mathrm{pH}\) of \(0.005\,\,{\text{M}}\,\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\) solution?
1 5.0
2 2.3
3 3.3
4 2.0
Explanation:
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) is dibasic acid, so \(0.005 \mathrm{M}=0.01 \mathrm{~N}\) \(\begin{gathered}=10^{-2} \mathrm{~N} \\\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\=-\log _{10}\left[10^{-2}\right]=2.0\end{gathered}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
315060
Calculate the \({\text{pH of 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{M Ca(OH}}{{\text{)}}_{\text{2}}}\). Given \(\log 2=0.3\)
315056
The number of hydroxyl ions in \(100 \mathrm{~mL}\) of a solution having \({\text{pH = }}10\) is
1 \(1 \times 10^{4}\)
2 \(3.012 \times 10^{4}\)
3 \(6.02 \times 10^{18}\)
4 \(6.023 \times 10^{19}\)
Explanation:
Given, \(\mathrm{pH}=10\) \(\therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10} \mathrm{~mol} / \mathrm{L}\) \(\therefore\left[\mathrm{OH}^{-}\right]=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(\because\) The number of hydroxyl ions present in 1 mole \(=6.023 \times 10^{23}\) \(\therefore\) The number of hydroxyl ion in \(1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(=6.023 \times 10^{23} \times 10^{-4} / \mathrm{L}\) \(=6.023 \times 10^{19} \mathrm{~mol} / \mathrm{L}\) \(\because\) In \(1000 \mathrm{~mL}\), number of hydroxyl ion \(=6.023 \times 10^{19}\) \(\therefore\) In \(100 \mathrm{~mL}\), number of hydroxyl ions \(=\dfrac{6.023 \times 10^{19} \times 100}{1000}=6.02 \times 10^{18}\)
AIIMS - 2009
CHXI07:EQUILIBRIUM
315057
What is the \({\text{pH}}{\mkern 1mu} \) of a \({\text{1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) \({\text{O}}{{\text{H}}^{\text{ - }}}\) solution at \(330 \mathrm{~K}\), if \({{\text{K}}_{\text{w}}}\) at \(330 \mathrm{~K}\) is \(10^{-13.6}\) ?
315058
How many litres of water must be added to 1 litre on an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ?
1 0.9 L
2 2.0 L
3 9.0 L
4 0.1 L
Explanation:
\((\mathrm{pH})_{1}=1 \Rightarrow\left[\mathrm{H}^{+}\right]_{1}=10^{-1}\) \((\mathrm{pH})_{2}=2 \Rightarrow\left[\mathrm{H}^{+}\right]_{2}=10^{-2}\) Dilution law, \({{\rm{M}}_1}\;{{\rm{V}}_{\rm{1}}} = {{\rm{M}}_{\rm{2}}}\;{{\rm{V}}_{\rm{2}}}\) \({10^{ - 1}} \times 1 = {10^{ - 2}} \times {{\rm{V}}_{\rm{2}}}\;\) \({{\rm{V}}_2} = 10\;{\rm{L}}\) Volume of water to be added \(=\mathrm{V}_{2}-\mathrm{V}_{1}=10^{-1}=9 \mathrm{~L}\) (on for every ten times dilution pH of strong acid increases by 1 unit).
CHXI07:EQUILIBRIUM
315059
What is the \(\mathrm{pH}\) of \(0.005\,\,{\text{M}}\,\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\) solution?
1 5.0
2 2.3
3 3.3
4 2.0
Explanation:
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) is dibasic acid, so \(0.005 \mathrm{M}=0.01 \mathrm{~N}\) \(\begin{gathered}=10^{-2} \mathrm{~N} \\\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\=-\log _{10}\left[10^{-2}\right]=2.0\end{gathered}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
315060
Calculate the \({\text{pH of 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{M Ca(OH}}{{\text{)}}_{\text{2}}}\). Given \(\log 2=0.3\)
315056
The number of hydroxyl ions in \(100 \mathrm{~mL}\) of a solution having \({\text{pH = }}10\) is
1 \(1 \times 10^{4}\)
2 \(3.012 \times 10^{4}\)
3 \(6.02 \times 10^{18}\)
4 \(6.023 \times 10^{19}\)
Explanation:
Given, \(\mathrm{pH}=10\) \(\therefore\left[\mathrm{H}^{+}\right]=1 \times 10^{-\mathrm{pH}}=1 \times 10^{-10} \mathrm{~mol} / \mathrm{L}\) \(\therefore\left[\mathrm{OH}^{-}\right]=\dfrac{1 \times 10^{-14}}{1 \times 10^{-10}}=1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(\because\) The number of hydroxyl ions present in 1 mole \(=6.023 \times 10^{23}\) \(\therefore\) The number of hydroxyl ion in \(1 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\) \(=6.023 \times 10^{23} \times 10^{-4} / \mathrm{L}\) \(=6.023 \times 10^{19} \mathrm{~mol} / \mathrm{L}\) \(\because\) In \(1000 \mathrm{~mL}\), number of hydroxyl ion \(=6.023 \times 10^{19}\) \(\therefore\) In \(100 \mathrm{~mL}\), number of hydroxyl ions \(=\dfrac{6.023 \times 10^{19} \times 100}{1000}=6.02 \times 10^{18}\)
AIIMS - 2009
CHXI07:EQUILIBRIUM
315057
What is the \({\text{pH}}{\mkern 1mu} \) of a \({\text{1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) \({\text{O}}{{\text{H}}^{\text{ - }}}\) solution at \(330 \mathrm{~K}\), if \({{\text{K}}_{\text{w}}}\) at \(330 \mathrm{~K}\) is \(10^{-13.6}\) ?
315058
How many litres of water must be added to 1 litre on an aqueous solution of HCl with a pH of 1 to create an aqueous solution with pH of 2 ?
1 0.9 L
2 2.0 L
3 9.0 L
4 0.1 L
Explanation:
\((\mathrm{pH})_{1}=1 \Rightarrow\left[\mathrm{H}^{+}\right]_{1}=10^{-1}\) \((\mathrm{pH})_{2}=2 \Rightarrow\left[\mathrm{H}^{+}\right]_{2}=10^{-2}\) Dilution law, \({{\rm{M}}_1}\;{{\rm{V}}_{\rm{1}}} = {{\rm{M}}_{\rm{2}}}\;{{\rm{V}}_{\rm{2}}}\) \({10^{ - 1}} \times 1 = {10^{ - 2}} \times {{\rm{V}}_{\rm{2}}}\;\) \({{\rm{V}}_2} = 10\;{\rm{L}}\) Volume of water to be added \(=\mathrm{V}_{2}-\mathrm{V}_{1}=10^{-1}=9 \mathrm{~L}\) (on for every ten times dilution pH of strong acid increases by 1 unit).
CHXI07:EQUILIBRIUM
315059
What is the \(\mathrm{pH}\) of \(0.005\,\,{\text{M}}\,\,{{\text{H}}_2}{\text{S}}{{\text{O}}_4}\) solution?
1 5.0
2 2.3
3 3.3
4 2.0
Explanation:
\(\mathrm{H}_{2} \mathrm{SO}_{4}\) is dibasic acid, so \(0.005 \mathrm{M}=0.01 \mathrm{~N}\) \(\begin{gathered}=10^{-2} \mathrm{~N} \\\mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\=-\log _{10}\left[10^{-2}\right]=2.0\end{gathered}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
315060
Calculate the \({\text{pH of 1}}{{\text{0}}^{{\text{ - 2}}}}{\text{M Ca(OH}}{{\text{)}}_{\text{2}}}\). Given \(\log 2=0.3\)
