315024
One litre of an aqueous solution has 3.65 g of HCl . If it is desired to increase the pH of the solution to 2 then \(\mathrm{H}_{3} \mathrm{O}^{+}\)ion concentration present initially should be
315015
The \(\mathrm{pH}\) of the solution obtained by mixing \({\text{100 }}\) \(\mathrm{mL}\) of a solution of \(\mathrm{pH}=3\) with \(400 \mathrm{~mL}\) of a solution of \(\mathrm{pH}=4\) is
1 \(3-\log 2.8\)
2 \(7-\log 2.8\)
3 \(4-\log 2.8\)
4 \(5-\log 2.8\)
Explanation:
\({\text{pH = 3 means 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ M}}\) and \({\text{pH = 4 means 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\). \({\text{100}}\;{\text{mL }}of{\text{ 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ M}}\) solution contain \(\mathrm {=\dfrac{10^{-3} \times 100}{1000}=10 \times 10^{-5} \mathrm{~mol}}\) \({\text{400}}\;{\text{mL }}of{\text{ 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) solution contain \(\mathrm {=\dfrac{10^{-4} \times 400}{1000}=4 \times 10^{-5} \mathrm{~mol}}\) Total number of moles present \(\mathrm {=\left(10 \times 10^{-5}\right)+\left(4 \times 10^{-5}\right)=14 \times 10^{-5} \mathrm{~mol}}\) Total volume after mixing \(\mathrm {=(100+400)=500}\) \(\mathrm {\mathrm{mL}}\) Molarity of the mixture \(\mathrm {=\dfrac{14 \times 10^{-5}}{500} \times 1000}\) \(\rm {=\left(\dfrac{14}{5} \times 10^{-4}\right) \mathrm{M}}\) \(\mathrm {\mathrm{pH}}\) of the mixture solution \(\mathrm {=-\log \left(\dfrac{14}{5} \times 10^{-4}\right)}\) \(\mathrm {\mathrm{pH}=4-\log 2.8}\) \(\rm {=-\log 2.8-(-4) \Rightarrow 4-\log 2.8}\)
KCET - 2011
CHXI07:EQUILIBRIUM
315016
What is the \(\mathrm{pH}\) value of \(10^{-2}\) molar \(\mathrm{HNO}_{3}\) solution?}
315024
One litre of an aqueous solution has 3.65 g of HCl . If it is desired to increase the pH of the solution to 2 then \(\mathrm{H}_{3} \mathrm{O}^{+}\)ion concentration present initially should be
315015
The \(\mathrm{pH}\) of the solution obtained by mixing \({\text{100 }}\) \(\mathrm{mL}\) of a solution of \(\mathrm{pH}=3\) with \(400 \mathrm{~mL}\) of a solution of \(\mathrm{pH}=4\) is
1 \(3-\log 2.8\)
2 \(7-\log 2.8\)
3 \(4-\log 2.8\)
4 \(5-\log 2.8\)
Explanation:
\({\text{pH = 3 means 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ M}}\) and \({\text{pH = 4 means 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\). \({\text{100}}\;{\text{mL }}of{\text{ 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ M}}\) solution contain \(\mathrm {=\dfrac{10^{-3} \times 100}{1000}=10 \times 10^{-5} \mathrm{~mol}}\) \({\text{400}}\;{\text{mL }}of{\text{ 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) solution contain \(\mathrm {=\dfrac{10^{-4} \times 400}{1000}=4 \times 10^{-5} \mathrm{~mol}}\) Total number of moles present \(\mathrm {=\left(10 \times 10^{-5}\right)+\left(4 \times 10^{-5}\right)=14 \times 10^{-5} \mathrm{~mol}}\) Total volume after mixing \(\mathrm {=(100+400)=500}\) \(\mathrm {\mathrm{mL}}\) Molarity of the mixture \(\mathrm {=\dfrac{14 \times 10^{-5}}{500} \times 1000}\) \(\rm {=\left(\dfrac{14}{5} \times 10^{-4}\right) \mathrm{M}}\) \(\mathrm {\mathrm{pH}}\) of the mixture solution \(\mathrm {=-\log \left(\dfrac{14}{5} \times 10^{-4}\right)}\) \(\mathrm {\mathrm{pH}=4-\log 2.8}\) \(\rm {=-\log 2.8-(-4) \Rightarrow 4-\log 2.8}\)
KCET - 2011
CHXI07:EQUILIBRIUM
315016
What is the \(\mathrm{pH}\) value of \(10^{-2}\) molar \(\mathrm{HNO}_{3}\) solution?}
315024
One litre of an aqueous solution has 3.65 g of HCl . If it is desired to increase the pH of the solution to 2 then \(\mathrm{H}_{3} \mathrm{O}^{+}\)ion concentration present initially should be
315015
The \(\mathrm{pH}\) of the solution obtained by mixing \({\text{100 }}\) \(\mathrm{mL}\) of a solution of \(\mathrm{pH}=3\) with \(400 \mathrm{~mL}\) of a solution of \(\mathrm{pH}=4\) is
1 \(3-\log 2.8\)
2 \(7-\log 2.8\)
3 \(4-\log 2.8\)
4 \(5-\log 2.8\)
Explanation:
\({\text{pH = 3 means 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ M}}\) and \({\text{pH = 4 means 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\). \({\text{100}}\;{\text{mL }}of{\text{ 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ M}}\) solution contain \(\mathrm {=\dfrac{10^{-3} \times 100}{1000}=10 \times 10^{-5} \mathrm{~mol}}\) \({\text{400}}\;{\text{mL }}of{\text{ 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) solution contain \(\mathrm {=\dfrac{10^{-4} \times 400}{1000}=4 \times 10^{-5} \mathrm{~mol}}\) Total number of moles present \(\mathrm {=\left(10 \times 10^{-5}\right)+\left(4 \times 10^{-5}\right)=14 \times 10^{-5} \mathrm{~mol}}\) Total volume after mixing \(\mathrm {=(100+400)=500}\) \(\mathrm {\mathrm{mL}}\) Molarity of the mixture \(\mathrm {=\dfrac{14 \times 10^{-5}}{500} \times 1000}\) \(\rm {=\left(\dfrac{14}{5} \times 10^{-4}\right) \mathrm{M}}\) \(\mathrm {\mathrm{pH}}\) of the mixture solution \(\mathrm {=-\log \left(\dfrac{14}{5} \times 10^{-4}\right)}\) \(\mathrm {\mathrm{pH}=4-\log 2.8}\) \(\rm {=-\log 2.8-(-4) \Rightarrow 4-\log 2.8}\)
KCET - 2011
CHXI07:EQUILIBRIUM
315016
What is the \(\mathrm{pH}\) value of \(10^{-2}\) molar \(\mathrm{HNO}_{3}\) solution?}
315024
One litre of an aqueous solution has 3.65 g of HCl . If it is desired to increase the pH of the solution to 2 then \(\mathrm{H}_{3} \mathrm{O}^{+}\)ion concentration present initially should be
315015
The \(\mathrm{pH}\) of the solution obtained by mixing \({\text{100 }}\) \(\mathrm{mL}\) of a solution of \(\mathrm{pH}=3\) with \(400 \mathrm{~mL}\) of a solution of \(\mathrm{pH}=4\) is
1 \(3-\log 2.8\)
2 \(7-\log 2.8\)
3 \(4-\log 2.8\)
4 \(5-\log 2.8\)
Explanation:
\({\text{pH = 3 means 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ M}}\) and \({\text{pH = 4 means 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\). \({\text{100}}\;{\text{mL }}of{\text{ 1}}{{\text{0}}^{{\text{ - 3}}}}{\text{ M}}\) solution contain \(\mathrm {=\dfrac{10^{-3} \times 100}{1000}=10 \times 10^{-5} \mathrm{~mol}}\) \({\text{400}}\;{\text{mL }}of{\text{ 1}}{{\text{0}}^{{\text{ - 4}}}}{\text{ M}}\) solution contain \(\mathrm {=\dfrac{10^{-4} \times 400}{1000}=4 \times 10^{-5} \mathrm{~mol}}\) Total number of moles present \(\mathrm {=\left(10 \times 10^{-5}\right)+\left(4 \times 10^{-5}\right)=14 \times 10^{-5} \mathrm{~mol}}\) Total volume after mixing \(\mathrm {=(100+400)=500}\) \(\mathrm {\mathrm{mL}}\) Molarity of the mixture \(\mathrm {=\dfrac{14 \times 10^{-5}}{500} \times 1000}\) \(\rm {=\left(\dfrac{14}{5} \times 10^{-4}\right) \mathrm{M}}\) \(\mathrm {\mathrm{pH}}\) of the mixture solution \(\mathrm {=-\log \left(\dfrac{14}{5} \times 10^{-4}\right)}\) \(\mathrm {\mathrm{pH}=4-\log 2.8}\) \(\rm {=-\log 2.8-(-4) \Rightarrow 4-\log 2.8}\)
KCET - 2011
CHXI07:EQUILIBRIUM
315016
What is the \(\mathrm{pH}\) value of \(10^{-2}\) molar \(\mathrm{HNO}_{3}\) solution?}
