314751
Aqueous solution of \(\mathrm{KCl}\) is neutral because
1 \(\mathrm{K}^{+}\)undergoes hydrolysis
2 \(\mathrm{Cl}^{-}\)undergoes hydrolysis
3 Both \(\mathrm{K}^{+}\)and \(\mathrm{Cl}^{-}\)undergo hydrolysis
4 No hydrolysis takes place'
Explanation:
Aqueous solution of \(\mathrm {\mathrm{KCl}}\) is neutral because it is made up of strong acid and strong base.
CHXI07:EQUILIBRIUM
314752
What is the \(\mathrm{pH}\) of \({\text{0}}{\text{.10 M}}{\mkern 1mu} {\mkern 1mu} {\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}\) solution? Hydrolysis constant of sodium acetate is \(5.6 \times 10^{-10}\)
314753
Which among the following salts undergoes hydrolysis?
1 \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)
2 \(\mathrm{KCl}\)
3 \(\mathrm{NH}_{4} \mathrm{Cl}\)
4 \(\mathrm{KNO}_{3}\)
Explanation:
\(\mathrm{NH}_{4} \mathrm{Cl}\) is a salt of weak base, \(\mathrm{NH}_{4} \mathrm{OH}\) and strong acid, \(\mathrm{HCl}\). Hence, it undergoes hydrolysis. While \(\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{KCl}\) and \(\mathrm{KNO}_{3}\) are salts of strong acids and strong bases and does not undergo hydrolysis. Note: Salts of SA and SB do not undergo hydrolysis.
MHTCET - 2021
CHXI07:EQUILIBRIUM
314754
Assuming that the degree of hydrolysis is small, the \(\mathrm{pH}\) of \({\text{0}}{\text{.1}}\,\,{\text{M}}\) solution of sodium acetate \(\left( {{{\text{K}}_{\text{a}}}{\text{ = 1}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}} \right)\) will be:
1 5.0
2 6.0
3 8.0
4 9.0
Explanation:
Sodium acetate is a salt of strong base and weak acid. \(\mathrm {\therefore \rm {\,{\text{pH}}\,}=7+\dfrac{1}{2} p K_{a}+\dfrac{1}{2} \log c}\) where \({\text{P}}{{\text{K}}_{\text{a}}}{\text{ = - log}}{{\text{K}}_{\text{a}}}\) \[{\text{ = 7 + }}\frac{{\text{5}}}{{\text{2}}}{\text{ - }}\frac{{\text{1}}}{{\text{2}}}\;\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{c}} {\therefore \;\;\;{\mkern 1mu} {\kern 1pt} {\text{ = - log1}}{{\text{0}}^{{\text{ - 5}}}}{\text{ = 5}}} \\ {{\text{logc = log1}}{{\text{0}}^{{\text{ - 1}}}}{\text{ = - 1}}} \end{array}} \right\}{\text{ = 9}}{\text{.0}}\]
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI07:EQUILIBRIUM
314751
Aqueous solution of \(\mathrm{KCl}\) is neutral because
1 \(\mathrm{K}^{+}\)undergoes hydrolysis
2 \(\mathrm{Cl}^{-}\)undergoes hydrolysis
3 Both \(\mathrm{K}^{+}\)and \(\mathrm{Cl}^{-}\)undergo hydrolysis
4 No hydrolysis takes place'
Explanation:
Aqueous solution of \(\mathrm {\mathrm{KCl}}\) is neutral because it is made up of strong acid and strong base.
CHXI07:EQUILIBRIUM
314752
What is the \(\mathrm{pH}\) of \({\text{0}}{\text{.10 M}}{\mkern 1mu} {\mkern 1mu} {\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}\) solution? Hydrolysis constant of sodium acetate is \(5.6 \times 10^{-10}\)
314753
Which among the following salts undergoes hydrolysis?
1 \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)
2 \(\mathrm{KCl}\)
3 \(\mathrm{NH}_{4} \mathrm{Cl}\)
4 \(\mathrm{KNO}_{3}\)
Explanation:
\(\mathrm{NH}_{4} \mathrm{Cl}\) is a salt of weak base, \(\mathrm{NH}_{4} \mathrm{OH}\) and strong acid, \(\mathrm{HCl}\). Hence, it undergoes hydrolysis. While \(\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{KCl}\) and \(\mathrm{KNO}_{3}\) are salts of strong acids and strong bases and does not undergo hydrolysis. Note: Salts of SA and SB do not undergo hydrolysis.
MHTCET - 2021
CHXI07:EQUILIBRIUM
314754
Assuming that the degree of hydrolysis is small, the \(\mathrm{pH}\) of \({\text{0}}{\text{.1}}\,\,{\text{M}}\) solution of sodium acetate \(\left( {{{\text{K}}_{\text{a}}}{\text{ = 1}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}} \right)\) will be:
1 5.0
2 6.0
3 8.0
4 9.0
Explanation:
Sodium acetate is a salt of strong base and weak acid. \(\mathrm {\therefore \rm {\,{\text{pH}}\,}=7+\dfrac{1}{2} p K_{a}+\dfrac{1}{2} \log c}\) where \({\text{P}}{{\text{K}}_{\text{a}}}{\text{ = - log}}{{\text{K}}_{\text{a}}}\) \[{\text{ = 7 + }}\frac{{\text{5}}}{{\text{2}}}{\text{ - }}\frac{{\text{1}}}{{\text{2}}}\;\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{c}} {\therefore \;\;\;{\mkern 1mu} {\kern 1pt} {\text{ = - log1}}{{\text{0}}^{{\text{ - 5}}}}{\text{ = 5}}} \\ {{\text{logc = log1}}{{\text{0}}^{{\text{ - 1}}}}{\text{ = - 1}}} \end{array}} \right\}{\text{ = 9}}{\text{.0}}\]
314751
Aqueous solution of \(\mathrm{KCl}\) is neutral because
1 \(\mathrm{K}^{+}\)undergoes hydrolysis
2 \(\mathrm{Cl}^{-}\)undergoes hydrolysis
3 Both \(\mathrm{K}^{+}\)and \(\mathrm{Cl}^{-}\)undergo hydrolysis
4 No hydrolysis takes place'
Explanation:
Aqueous solution of \(\mathrm {\mathrm{KCl}}\) is neutral because it is made up of strong acid and strong base.
CHXI07:EQUILIBRIUM
314752
What is the \(\mathrm{pH}\) of \({\text{0}}{\text{.10 M}}{\mkern 1mu} {\mkern 1mu} {\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}\) solution? Hydrolysis constant of sodium acetate is \(5.6 \times 10^{-10}\)
314753
Which among the following salts undergoes hydrolysis?
1 \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)
2 \(\mathrm{KCl}\)
3 \(\mathrm{NH}_{4} \mathrm{Cl}\)
4 \(\mathrm{KNO}_{3}\)
Explanation:
\(\mathrm{NH}_{4} \mathrm{Cl}\) is a salt of weak base, \(\mathrm{NH}_{4} \mathrm{OH}\) and strong acid, \(\mathrm{HCl}\). Hence, it undergoes hydrolysis. While \(\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{KCl}\) and \(\mathrm{KNO}_{3}\) are salts of strong acids and strong bases and does not undergo hydrolysis. Note: Salts of SA and SB do not undergo hydrolysis.
MHTCET - 2021
CHXI07:EQUILIBRIUM
314754
Assuming that the degree of hydrolysis is small, the \(\mathrm{pH}\) of \({\text{0}}{\text{.1}}\,\,{\text{M}}\) solution of sodium acetate \(\left( {{{\text{K}}_{\text{a}}}{\text{ = 1}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}} \right)\) will be:
1 5.0
2 6.0
3 8.0
4 9.0
Explanation:
Sodium acetate is a salt of strong base and weak acid. \(\mathrm {\therefore \rm {\,{\text{pH}}\,}=7+\dfrac{1}{2} p K_{a}+\dfrac{1}{2} \log c}\) where \({\text{P}}{{\text{K}}_{\text{a}}}{\text{ = - log}}{{\text{K}}_{\text{a}}}\) \[{\text{ = 7 + }}\frac{{\text{5}}}{{\text{2}}}{\text{ - }}\frac{{\text{1}}}{{\text{2}}}\;\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{c}} {\therefore \;\;\;{\mkern 1mu} {\kern 1pt} {\text{ = - log1}}{{\text{0}}^{{\text{ - 5}}}}{\text{ = 5}}} \\ {{\text{logc = log1}}{{\text{0}}^{{\text{ - 1}}}}{\text{ = - 1}}} \end{array}} \right\}{\text{ = 9}}{\text{.0}}\]
314751
Aqueous solution of \(\mathrm{KCl}\) is neutral because
1 \(\mathrm{K}^{+}\)undergoes hydrolysis
2 \(\mathrm{Cl}^{-}\)undergoes hydrolysis
3 Both \(\mathrm{K}^{+}\)and \(\mathrm{Cl}^{-}\)undergo hydrolysis
4 No hydrolysis takes place'
Explanation:
Aqueous solution of \(\mathrm {\mathrm{KCl}}\) is neutral because it is made up of strong acid and strong base.
CHXI07:EQUILIBRIUM
314752
What is the \(\mathrm{pH}\) of \({\text{0}}{\text{.10 M}}{\mkern 1mu} {\mkern 1mu} {\text{C}}{{\text{H}}_{\text{3}}}{\text{COONa}}\) solution? Hydrolysis constant of sodium acetate is \(5.6 \times 10^{-10}\)
314753
Which among the following salts undergoes hydrolysis?
1 \(\mathrm{Na}_{2} \mathrm{SO}_{4}\)
2 \(\mathrm{KCl}\)
3 \(\mathrm{NH}_{4} \mathrm{Cl}\)
4 \(\mathrm{KNO}_{3}\)
Explanation:
\(\mathrm{NH}_{4} \mathrm{Cl}\) is a salt of weak base, \(\mathrm{NH}_{4} \mathrm{OH}\) and strong acid, \(\mathrm{HCl}\). Hence, it undergoes hydrolysis. While \(\mathrm{Na}_{2} \mathrm{SO}_{4}, \mathrm{KCl}\) and \(\mathrm{KNO}_{3}\) are salts of strong acids and strong bases and does not undergo hydrolysis. Note: Salts of SA and SB do not undergo hydrolysis.
MHTCET - 2021
CHXI07:EQUILIBRIUM
314754
Assuming that the degree of hydrolysis is small, the \(\mathrm{pH}\) of \({\text{0}}{\text{.1}}\,\,{\text{M}}\) solution of sodium acetate \(\left( {{{\text{K}}_{\text{a}}}{\text{ = 1}}{\text{.0}} \times {\text{1}}{{\text{0}}^{{\text{ - 5}}}}} \right)\) will be:
1 5.0
2 6.0
3 8.0
4 9.0
Explanation:
Sodium acetate is a salt of strong base and weak acid. \(\mathrm {\therefore \rm {\,{\text{pH}}\,}=7+\dfrac{1}{2} p K_{a}+\dfrac{1}{2} \log c}\) where \({\text{P}}{{\text{K}}_{\text{a}}}{\text{ = - log}}{{\text{K}}_{\text{a}}}\) \[{\text{ = 7 + }}\frac{{\text{5}}}{{\text{2}}}{\text{ - }}\frac{{\text{1}}}{{\text{2}}}\;\;\;{\mkern 1mu} {\kern 1pt} \left\{ {\begin{array}{*{20}{c}} {\therefore \;\;\;{\mkern 1mu} {\kern 1pt} {\text{ = - log1}}{{\text{0}}^{{\text{ - 5}}}}{\text{ = 5}}} \\ {{\text{logc = log1}}{{\text{0}}^{{\text{ - 1}}}}{\text{ = - 1}}} \end{array}} \right\}{\text{ = 9}}{\text{.0}}\]
