229671
The hydrolysis of sodium carbonate involves the reaction between
1 Sodium ion and water
2 \(\mathrm{Na}^{+}\)and \(\mathrm{OH}^{-}\)
3 \(\mathrm{CO}_3^{2-}\) and water
4 \(\mathrm{CO}_3^{2-}\) and \(\mathrm{H}^{+}\)
Explanation:
: Reactions of hydrolysis would be the opposite of condensation reaction in which two molecules bind together into a larger one an expel a molecule of water. Hydrolysis thus requires water to breakdown, while condensation occurs by eliminating water and all the other solvents. Any reactions to the hydration are hydrolysis. Now, we will write the reaction of hydrolysis of sodium carbonate- \(\begin{array}{ll} & \mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \quad 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{CO}_3 \\ \text { or } & 2 \mathrm{Na}^{+}+\mathrm{CO}_3^{2-}+2 \mathrm{H}_2 \mathrm{O} \quad 2 \mathrm{Na}^{+}+2 \mathrm{OH}^{-}+\mathrm{H}_2 \mathrm{CO}_3 \end{array}\) It is anionic is alkaline in nature due to hydrolysis.
MHT CET-2007
Ionic Equilibrium
229673
Conjugate base for Bronsted acids \(\mathrm{H}_2 \mathrm{O}\) and \(\mathrm{HF}\) are
: Bronsted acid- Conjugate base \(\begin{aligned} & \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{OH}^{-}+\mathrm{H}^{+} \\ & \mathrm{HF} \rightarrow \mathrm{F}^{-}+\mathrm{H}^{+} \end{aligned}\) Hence, \(\mathrm{OH}^{-}\)and \(\mathrm{F}^{-}\)are conjugate base of Bronsted acid \(\left(\mathrm{H}_2 \mathrm{O}\right)\) and \(\mathrm{HF}\).
NEET-2019
Ionic Equilibrium
229675
Which of the following fluoro-compound is most likely to behave as a Lewis base?
1 \(\mathrm{BF}_3\)
2 \(\mathrm{PF}_3\)
3 \(\mathrm{CF}_4\)
4 \(\mathrm{SiF}_4\)
Explanation:
: Lewis base is substance which contains lone pair on central atom. Phosphorus is a group VA element so it has 5 valence electron. Fluorine is a group VII A element so it has 7 valence electrons. So, total valence electrons \(=5+7(3)\) \(=5+21=26\) As there is a lone pair on the central phosphorus atom. So \(\mathrm{PF}_3\) tends to donate its lone pair of electrons and behave as Lewis base.
NEET-II 2016
Ionic Equilibrium
229689
Two acids \(A\) and \(B\) have \(\mathrm{pK}_{\mathrm{a}} 4\) and 6 , then
1 \(A\) is \(4 / 6\) times stronger than \(B\)
2 \(A\) is 10 times stronger than \(B\)
3 \(A\) is \(6 / 4\) times stronnger than \(B\)
4 \(\mathrm{B}\) is 10 times stronger than \(\mathrm{A}\)
Explanation:
\(\begin{aligned} & \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}} \\ & \therefore \quad\left(\mathrm{pK}_{\mathrm{a}}\right)_{\mathrm{A}}=-\log \left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{A}} \\ & \left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{A}}=\operatorname{antilog}(-4)=10^{-4} \\ & \end{aligned}\) Similarly, \(\left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{B}}=\operatorname{anti} \log (-6)=10^{-6}\) \(\begin{aligned} \frac{\text { Strength of } A}{\text { Strength of B }} & =\sqrt{\frac{K_a}{K_b}} \\ & =\sqrt{\frac{10^{-4}}{10^{-6}}}=10 \end{aligned}\) \(\therefore\) Strength of A \(=10 \times\) strength of B.
229671
The hydrolysis of sodium carbonate involves the reaction between
1 Sodium ion and water
2 \(\mathrm{Na}^{+}\)and \(\mathrm{OH}^{-}\)
3 \(\mathrm{CO}_3^{2-}\) and water
4 \(\mathrm{CO}_3^{2-}\) and \(\mathrm{H}^{+}\)
Explanation:
: Reactions of hydrolysis would be the opposite of condensation reaction in which two molecules bind together into a larger one an expel a molecule of water. Hydrolysis thus requires water to breakdown, while condensation occurs by eliminating water and all the other solvents. Any reactions to the hydration are hydrolysis. Now, we will write the reaction of hydrolysis of sodium carbonate- \(\begin{array}{ll} & \mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \quad 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{CO}_3 \\ \text { or } & 2 \mathrm{Na}^{+}+\mathrm{CO}_3^{2-}+2 \mathrm{H}_2 \mathrm{O} \quad 2 \mathrm{Na}^{+}+2 \mathrm{OH}^{-}+\mathrm{H}_2 \mathrm{CO}_3 \end{array}\) It is anionic is alkaline in nature due to hydrolysis.
MHT CET-2007
Ionic Equilibrium
229673
Conjugate base for Bronsted acids \(\mathrm{H}_2 \mathrm{O}\) and \(\mathrm{HF}\) are
: Bronsted acid- Conjugate base \(\begin{aligned} & \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{OH}^{-}+\mathrm{H}^{+} \\ & \mathrm{HF} \rightarrow \mathrm{F}^{-}+\mathrm{H}^{+} \end{aligned}\) Hence, \(\mathrm{OH}^{-}\)and \(\mathrm{F}^{-}\)are conjugate base of Bronsted acid \(\left(\mathrm{H}_2 \mathrm{O}\right)\) and \(\mathrm{HF}\).
NEET-2019
Ionic Equilibrium
229675
Which of the following fluoro-compound is most likely to behave as a Lewis base?
1 \(\mathrm{BF}_3\)
2 \(\mathrm{PF}_3\)
3 \(\mathrm{CF}_4\)
4 \(\mathrm{SiF}_4\)
Explanation:
: Lewis base is substance which contains lone pair on central atom. Phosphorus is a group VA element so it has 5 valence electron. Fluorine is a group VII A element so it has 7 valence electrons. So, total valence electrons \(=5+7(3)\) \(=5+21=26\) As there is a lone pair on the central phosphorus atom. So \(\mathrm{PF}_3\) tends to donate its lone pair of electrons and behave as Lewis base.
NEET-II 2016
Ionic Equilibrium
229689
Two acids \(A\) and \(B\) have \(\mathrm{pK}_{\mathrm{a}} 4\) and 6 , then
1 \(A\) is \(4 / 6\) times stronger than \(B\)
2 \(A\) is 10 times stronger than \(B\)
3 \(A\) is \(6 / 4\) times stronnger than \(B\)
4 \(\mathrm{B}\) is 10 times stronger than \(\mathrm{A}\)
Explanation:
\(\begin{aligned} & \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}} \\ & \therefore \quad\left(\mathrm{pK}_{\mathrm{a}}\right)_{\mathrm{A}}=-\log \left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{A}} \\ & \left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{A}}=\operatorname{antilog}(-4)=10^{-4} \\ & \end{aligned}\) Similarly, \(\left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{B}}=\operatorname{anti} \log (-6)=10^{-6}\) \(\begin{aligned} \frac{\text { Strength of } A}{\text { Strength of B }} & =\sqrt{\frac{K_a}{K_b}} \\ & =\sqrt{\frac{10^{-4}}{10^{-6}}}=10 \end{aligned}\) \(\therefore\) Strength of A \(=10 \times\) strength of B.
229671
The hydrolysis of sodium carbonate involves the reaction between
1 Sodium ion and water
2 \(\mathrm{Na}^{+}\)and \(\mathrm{OH}^{-}\)
3 \(\mathrm{CO}_3^{2-}\) and water
4 \(\mathrm{CO}_3^{2-}\) and \(\mathrm{H}^{+}\)
Explanation:
: Reactions of hydrolysis would be the opposite of condensation reaction in which two molecules bind together into a larger one an expel a molecule of water. Hydrolysis thus requires water to breakdown, while condensation occurs by eliminating water and all the other solvents. Any reactions to the hydration are hydrolysis. Now, we will write the reaction of hydrolysis of sodium carbonate- \(\begin{array}{ll} & \mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \quad 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{CO}_3 \\ \text { or } & 2 \mathrm{Na}^{+}+\mathrm{CO}_3^{2-}+2 \mathrm{H}_2 \mathrm{O} \quad 2 \mathrm{Na}^{+}+2 \mathrm{OH}^{-}+\mathrm{H}_2 \mathrm{CO}_3 \end{array}\) It is anionic is alkaline in nature due to hydrolysis.
MHT CET-2007
Ionic Equilibrium
229673
Conjugate base for Bronsted acids \(\mathrm{H}_2 \mathrm{O}\) and \(\mathrm{HF}\) are
: Bronsted acid- Conjugate base \(\begin{aligned} & \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{OH}^{-}+\mathrm{H}^{+} \\ & \mathrm{HF} \rightarrow \mathrm{F}^{-}+\mathrm{H}^{+} \end{aligned}\) Hence, \(\mathrm{OH}^{-}\)and \(\mathrm{F}^{-}\)are conjugate base of Bronsted acid \(\left(\mathrm{H}_2 \mathrm{O}\right)\) and \(\mathrm{HF}\).
NEET-2019
Ionic Equilibrium
229675
Which of the following fluoro-compound is most likely to behave as a Lewis base?
1 \(\mathrm{BF}_3\)
2 \(\mathrm{PF}_3\)
3 \(\mathrm{CF}_4\)
4 \(\mathrm{SiF}_4\)
Explanation:
: Lewis base is substance which contains lone pair on central atom. Phosphorus is a group VA element so it has 5 valence electron. Fluorine is a group VII A element so it has 7 valence electrons. So, total valence electrons \(=5+7(3)\) \(=5+21=26\) As there is a lone pair on the central phosphorus atom. So \(\mathrm{PF}_3\) tends to donate its lone pair of electrons and behave as Lewis base.
NEET-II 2016
Ionic Equilibrium
229689
Two acids \(A\) and \(B\) have \(\mathrm{pK}_{\mathrm{a}} 4\) and 6 , then
1 \(A\) is \(4 / 6\) times stronger than \(B\)
2 \(A\) is 10 times stronger than \(B\)
3 \(A\) is \(6 / 4\) times stronnger than \(B\)
4 \(\mathrm{B}\) is 10 times stronger than \(\mathrm{A}\)
Explanation:
\(\begin{aligned} & \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}} \\ & \therefore \quad\left(\mathrm{pK}_{\mathrm{a}}\right)_{\mathrm{A}}=-\log \left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{A}} \\ & \left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{A}}=\operatorname{antilog}(-4)=10^{-4} \\ & \end{aligned}\) Similarly, \(\left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{B}}=\operatorname{anti} \log (-6)=10^{-6}\) \(\begin{aligned} \frac{\text { Strength of } A}{\text { Strength of B }} & =\sqrt{\frac{K_a}{K_b}} \\ & =\sqrt{\frac{10^{-4}}{10^{-6}}}=10 \end{aligned}\) \(\therefore\) Strength of A \(=10 \times\) strength of B.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ionic Equilibrium
229671
The hydrolysis of sodium carbonate involves the reaction between
1 Sodium ion and water
2 \(\mathrm{Na}^{+}\)and \(\mathrm{OH}^{-}\)
3 \(\mathrm{CO}_3^{2-}\) and water
4 \(\mathrm{CO}_3^{2-}\) and \(\mathrm{H}^{+}\)
Explanation:
: Reactions of hydrolysis would be the opposite of condensation reaction in which two molecules bind together into a larger one an expel a molecule of water. Hydrolysis thus requires water to breakdown, while condensation occurs by eliminating water and all the other solvents. Any reactions to the hydration are hydrolysis. Now, we will write the reaction of hydrolysis of sodium carbonate- \(\begin{array}{ll} & \mathrm{Na}_2 \mathrm{CO}_3+2 \mathrm{H}_2 \mathrm{O} \quad 2 \mathrm{NaOH}+\mathrm{H}_2 \mathrm{CO}_3 \\ \text { or } & 2 \mathrm{Na}^{+}+\mathrm{CO}_3^{2-}+2 \mathrm{H}_2 \mathrm{O} \quad 2 \mathrm{Na}^{+}+2 \mathrm{OH}^{-}+\mathrm{H}_2 \mathrm{CO}_3 \end{array}\) It is anionic is alkaline in nature due to hydrolysis.
MHT CET-2007
Ionic Equilibrium
229673
Conjugate base for Bronsted acids \(\mathrm{H}_2 \mathrm{O}\) and \(\mathrm{HF}\) are
: Bronsted acid- Conjugate base \(\begin{aligned} & \mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{OH}^{-}+\mathrm{H}^{+} \\ & \mathrm{HF} \rightarrow \mathrm{F}^{-}+\mathrm{H}^{+} \end{aligned}\) Hence, \(\mathrm{OH}^{-}\)and \(\mathrm{F}^{-}\)are conjugate base of Bronsted acid \(\left(\mathrm{H}_2 \mathrm{O}\right)\) and \(\mathrm{HF}\).
NEET-2019
Ionic Equilibrium
229675
Which of the following fluoro-compound is most likely to behave as a Lewis base?
1 \(\mathrm{BF}_3\)
2 \(\mathrm{PF}_3\)
3 \(\mathrm{CF}_4\)
4 \(\mathrm{SiF}_4\)
Explanation:
: Lewis base is substance which contains lone pair on central atom. Phosphorus is a group VA element so it has 5 valence electron. Fluorine is a group VII A element so it has 7 valence electrons. So, total valence electrons \(=5+7(3)\) \(=5+21=26\) As there is a lone pair on the central phosphorus atom. So \(\mathrm{PF}_3\) tends to donate its lone pair of electrons and behave as Lewis base.
NEET-II 2016
Ionic Equilibrium
229689
Two acids \(A\) and \(B\) have \(\mathrm{pK}_{\mathrm{a}} 4\) and 6 , then
1 \(A\) is \(4 / 6\) times stronger than \(B\)
2 \(A\) is 10 times stronger than \(B\)
3 \(A\) is \(6 / 4\) times stronnger than \(B\)
4 \(\mathrm{B}\) is 10 times stronger than \(\mathrm{A}\)
Explanation:
\(\begin{aligned} & \mathrm{pK}_{\mathrm{a}}=-\log \mathrm{K}_{\mathrm{a}} \\ & \therefore \quad\left(\mathrm{pK}_{\mathrm{a}}\right)_{\mathrm{A}}=-\log \left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{A}} \\ & \left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{A}}=\operatorname{antilog}(-4)=10^{-4} \\ & \end{aligned}\) Similarly, \(\left(\mathrm{K}_{\mathrm{a}}\right)_{\mathrm{B}}=\operatorname{anti} \log (-6)=10^{-6}\) \(\begin{aligned} \frac{\text { Strength of } A}{\text { Strength of B }} & =\sqrt{\frac{K_a}{K_b}} \\ & =\sqrt{\frac{10^{-4}}{10^{-6}}}=10 \end{aligned}\) \(\therefore\) Strength of A \(=10 \times\) strength of B.
