315052
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? [Given: \({\mathrm{\log _{10}(45)=1.6532}}\) ]
1 12.65
2 7
3 1.35
4 14
Explanation:
Assuming that 1000 mL of 0.1 M NaOH is mixed with 1000 mL of 0.01 M HCl . \(\mathop {{\text{NaOH}}}\limits_{{\text{0}}{\text{.1}}\;{\text{mol}}} {\text{ + }}\mathop {{\text{HCl}}}\limits_{{\text{0}}{\text{.01}}\;{\text{mol}}} \to \mathop {{\text{NaCl}}}\limits_{{\text{0}}{\text{.01}}\;{\text{mol}}} {\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\) Here, NaOH is the excess reagent. Hence, the pH of the solution will be greater than 7.0 . Excess amount of \({\mathrm{\mathrm{NaOH}=0.09 \mathrm{~mol}}}\) The volume of the resulting solution \({\mathrm{\approx 2000 \mathrm{~mL} \approx 2 \mathrm{~L}}}\) \({\mathrm{\left[\mathrm{OH}^{-}\right]=\dfrac{0.09}{2}=0.045 \mathrm{~mol} / \mathrm{L}}}\) \({\text{pOH }} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]\) \( = - {\log _{10}}\left( {45 \times {{10}^{ - 3}}} \right)\) \( = - [1.6532 - 3] = 1.3468\) \({\text{pH }} + {\text{pOH}} = 14\) \(\therefore \quad {\text{pH }} = 14 - {\text{pOH}} = 14 - 1.3468\) \( = 12.6532 \approx 12.65\)
CHXI07:EQUILIBRIUM
315053
What is the \(\mathrm{pH}\) of millimolar solution of \(\mathrm{NaOH}\) ?
1 13
2 11
3 3
4 2
Explanation:
One millimolar solution of \(\mathrm{NaOH}=1 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) of \(\mathrm{NaOH}\) \(\therefore\left[\mathrm{OH}^{-}\right]=1 \times 10^{-3} \mathrm{~mole} / \mathrm{L}\) \(\therefore \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(1 \times 10^{-3}\right)=3\) \(\therefore \mathrm{pH}=14-\mathrm{pOH}=14-3=11\)
\(\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{of 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{M NaOH = 1}}{{\text{0}}^{{\text{ - 6}}}}\) when 100 times diluted, \(\rm {\left[\mathrm{OH}^{-}\right]=\dfrac{10^{-6}}{10^{2}}=10^{-8}}\) As it is a weak base, autoprotolysis of water will also take place. So, the contribution of \(\mathrm {\mathrm{OH}^{-}}\) ions from water must also be considered. \(\mathrm {\therefore}\) Total \(\mathrm {\left[\mathrm{OH}^{-}\right]=10^{-8}+\left[\mathrm{OH}^{-}\right]_{\text {water }}}\)
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CHXI07:EQUILIBRIUM
315052
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? [Given: \({\mathrm{\log _{10}(45)=1.6532}}\) ]
1 12.65
2 7
3 1.35
4 14
Explanation:
Assuming that 1000 mL of 0.1 M NaOH is mixed with 1000 mL of 0.01 M HCl . \(\mathop {{\text{NaOH}}}\limits_{{\text{0}}{\text{.1}}\;{\text{mol}}} {\text{ + }}\mathop {{\text{HCl}}}\limits_{{\text{0}}{\text{.01}}\;{\text{mol}}} \to \mathop {{\text{NaCl}}}\limits_{{\text{0}}{\text{.01}}\;{\text{mol}}} {\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\) Here, NaOH is the excess reagent. Hence, the pH of the solution will be greater than 7.0 . Excess amount of \({\mathrm{\mathrm{NaOH}=0.09 \mathrm{~mol}}}\) The volume of the resulting solution \({\mathrm{\approx 2000 \mathrm{~mL} \approx 2 \mathrm{~L}}}\) \({\mathrm{\left[\mathrm{OH}^{-}\right]=\dfrac{0.09}{2}=0.045 \mathrm{~mol} / \mathrm{L}}}\) \({\text{pOH }} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]\) \( = - {\log _{10}}\left( {45 \times {{10}^{ - 3}}} \right)\) \( = - [1.6532 - 3] = 1.3468\) \({\text{pH }} + {\text{pOH}} = 14\) \(\therefore \quad {\text{pH }} = 14 - {\text{pOH}} = 14 - 1.3468\) \( = 12.6532 \approx 12.65\)
CHXI07:EQUILIBRIUM
315053
What is the \(\mathrm{pH}\) of millimolar solution of \(\mathrm{NaOH}\) ?
1 13
2 11
3 3
4 2
Explanation:
One millimolar solution of \(\mathrm{NaOH}=1 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) of \(\mathrm{NaOH}\) \(\therefore\left[\mathrm{OH}^{-}\right]=1 \times 10^{-3} \mathrm{~mole} / \mathrm{L}\) \(\therefore \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(1 \times 10^{-3}\right)=3\) \(\therefore \mathrm{pH}=14-\mathrm{pOH}=14-3=11\)
\(\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{of 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{M NaOH = 1}}{{\text{0}}^{{\text{ - 6}}}}\) when 100 times diluted, \(\rm {\left[\mathrm{OH}^{-}\right]=\dfrac{10^{-6}}{10^{2}}=10^{-8}}\) As it is a weak base, autoprotolysis of water will also take place. So, the contribution of \(\mathrm {\mathrm{OH}^{-}}\) ions from water must also be considered. \(\mathrm {\therefore}\) Total \(\mathrm {\left[\mathrm{OH}^{-}\right]=10^{-8}+\left[\mathrm{OH}^{-}\right]_{\text {water }}}\)
315052
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? [Given: \({\mathrm{\log _{10}(45)=1.6532}}\) ]
1 12.65
2 7
3 1.35
4 14
Explanation:
Assuming that 1000 mL of 0.1 M NaOH is mixed with 1000 mL of 0.01 M HCl . \(\mathop {{\text{NaOH}}}\limits_{{\text{0}}{\text{.1}}\;{\text{mol}}} {\text{ + }}\mathop {{\text{HCl}}}\limits_{{\text{0}}{\text{.01}}\;{\text{mol}}} \to \mathop {{\text{NaCl}}}\limits_{{\text{0}}{\text{.01}}\;{\text{mol}}} {\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\) Here, NaOH is the excess reagent. Hence, the pH of the solution will be greater than 7.0 . Excess amount of \({\mathrm{\mathrm{NaOH}=0.09 \mathrm{~mol}}}\) The volume of the resulting solution \({\mathrm{\approx 2000 \mathrm{~mL} \approx 2 \mathrm{~L}}}\) \({\mathrm{\left[\mathrm{OH}^{-}\right]=\dfrac{0.09}{2}=0.045 \mathrm{~mol} / \mathrm{L}}}\) \({\text{pOH }} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]\) \( = - {\log _{10}}\left( {45 \times {{10}^{ - 3}}} \right)\) \( = - [1.6532 - 3] = 1.3468\) \({\text{pH }} + {\text{pOH}} = 14\) \(\therefore \quad {\text{pH }} = 14 - {\text{pOH}} = 14 - 1.3468\) \( = 12.6532 \approx 12.65\)
CHXI07:EQUILIBRIUM
315053
What is the \(\mathrm{pH}\) of millimolar solution of \(\mathrm{NaOH}\) ?
1 13
2 11
3 3
4 2
Explanation:
One millimolar solution of \(\mathrm{NaOH}=1 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) of \(\mathrm{NaOH}\) \(\therefore\left[\mathrm{OH}^{-}\right]=1 \times 10^{-3} \mathrm{~mole} / \mathrm{L}\) \(\therefore \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(1 \times 10^{-3}\right)=3\) \(\therefore \mathrm{pH}=14-\mathrm{pOH}=14-3=11\)
\(\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{of 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{M NaOH = 1}}{{\text{0}}^{{\text{ - 6}}}}\) when 100 times diluted, \(\rm {\left[\mathrm{OH}^{-}\right]=\dfrac{10^{-6}}{10^{2}}=10^{-8}}\) As it is a weak base, autoprotolysis of water will also take place. So, the contribution of \(\mathrm {\mathrm{OH}^{-}}\) ions from water must also be considered. \(\mathrm {\therefore}\) Total \(\mathrm {\left[\mathrm{OH}^{-}\right]=10^{-8}+\left[\mathrm{OH}^{-}\right]_{\text {water }}}\)
315052
What is the pH of the resulting solution when equal volumes of 0.1 M NaOH and 0.01 M HCl are mixed? [Given: \({\mathrm{\log _{10}(45)=1.6532}}\) ]
1 12.65
2 7
3 1.35
4 14
Explanation:
Assuming that 1000 mL of 0.1 M NaOH is mixed with 1000 mL of 0.01 M HCl . \(\mathop {{\text{NaOH}}}\limits_{{\text{0}}{\text{.1}}\;{\text{mol}}} {\text{ + }}\mathop {{\text{HCl}}}\limits_{{\text{0}}{\text{.01}}\;{\text{mol}}} \to \mathop {{\text{NaCl}}}\limits_{{\text{0}}{\text{.01}}\;{\text{mol}}} {\text{ + }}{{\text{H}}_{\text{2}}}{\text{O}}\) Here, NaOH is the excess reagent. Hence, the pH of the solution will be greater than 7.0 . Excess amount of \({\mathrm{\mathrm{NaOH}=0.09 \mathrm{~mol}}}\) The volume of the resulting solution \({\mathrm{\approx 2000 \mathrm{~mL} \approx 2 \mathrm{~L}}}\) \({\mathrm{\left[\mathrm{OH}^{-}\right]=\dfrac{0.09}{2}=0.045 \mathrm{~mol} / \mathrm{L}}}\) \({\text{pOH }} = - {\log _{10}}\left[ {{\text{O}}{{\text{H}}^ - }} \right]\) \( = - {\log _{10}}\left( {45 \times {{10}^{ - 3}}} \right)\) \( = - [1.6532 - 3] = 1.3468\) \({\text{pH }} + {\text{pOH}} = 14\) \(\therefore \quad {\text{pH }} = 14 - {\text{pOH}} = 14 - 1.3468\) \( = 12.6532 \approx 12.65\)
CHXI07:EQUILIBRIUM
315053
What is the \(\mathrm{pH}\) of millimolar solution of \(\mathrm{NaOH}\) ?
1 13
2 11
3 3
4 2
Explanation:
One millimolar solution of \(\mathrm{NaOH}=1 \times 10^{-3} \mathrm{~mol} / \mathrm{L}\) of \(\mathrm{NaOH}\) \(\therefore\left[\mathrm{OH}^{-}\right]=1 \times 10^{-3} \mathrm{~mole} / \mathrm{L}\) \(\therefore \mathrm{pOH}=-\log \left[\mathrm{OH}^{-}\right]=-\log \left(1 \times 10^{-3}\right)=3\) \(\therefore \mathrm{pH}=14-\mathrm{pOH}=14-3=11\)
\(\left[ {{\text{O}}{{\text{H}}^{\text{ - }}}} \right]{\text{of 1}}{{\text{0}}^{{\text{ - 6}}}}{\text{M NaOH = 1}}{{\text{0}}^{{\text{ - 6}}}}\) when 100 times diluted, \(\rm {\left[\mathrm{OH}^{-}\right]=\dfrac{10^{-6}}{10^{2}}=10^{-8}}\) As it is a weak base, autoprotolysis of water will also take place. So, the contribution of \(\mathrm {\mathrm{OH}^{-}}\) ions from water must also be considered. \(\mathrm {\therefore}\) Total \(\mathrm {\left[\mathrm{OH}^{-}\right]=10^{-8}+\left[\mathrm{OH}^{-}\right]_{\text {water }}}\)
