230034
When common salt is added to a saturated solution of soap, soap is precipitated. This is based on the principle of
1 common ion effect
2 solubility product
3 adsorption form solution
4 peptisation.
Explanation:
The saturated solution of soap (RCOONa), show precipitation addition of NaCl , because an increase in \(\mathrm{Na}^{+}\)concentration helps in crossing over ionic product to their \(\mathrm{K}_{\mathrm{sp}}\) values. \[ \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}(\mathrm{~s}) \square \quad \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq}) \] When NaCl is added, \(\left[\mathrm{Na}^{+}\right]\)increases due to which ionic product, \(\left[\mathrm{RCOO}^{-}\right]\left[\mathrm{Na}^{+}\right]\)also increases which results in precipitation of soap. It can also be explained on the basis of common ion effect. Among the given compound the \(\mathrm{NH}_3\) is the most basic. Hence has highest proton affinity.
COMEDK-2018
Ionic Equilibrium
230035
Assertion (A) : \(\mathrm{Sb}_2 \mathrm{~S}_3\) is not soluble in yellow ammonium sulphide. Reason (R) : The common ion effect due to \(\mathrm{S}^{2-}\) ions reduces the stability of \(\mathbf{S b}_2 \mathrm{~S}_3\).
1 Both A and R are true and R is the correct explanation of the A .
2 Both A and R are true but R is not the correct explanation of the \(A\).
3 A is true, but R is false.
4 Both A and R are false statements.
Explanation:
(A) \(\mathrm{Sb}_2 \mathrm{~S}_3\) is soluble in yellow ammonium sulphide. \[ \mathrm{Sb}_2 \mathrm{~S}_3+3\left(\mathrm{NH}_4\right)_2 \mathrm{~S} \longrightarrow 2\left(\mathrm{NH}_4\right)_3 \mathrm{SbS}_3 \] Hence, the statement is false. (R) This is due to the fact that when the colour changes from white to yellow the number of sulphur increases. therefore the ions of the sulphur atoms increases i.e. \(\mathrm{S}^{2-}\) ions. So, as the amount of the sulphur ions increases the common ion effect also increases. Therefore, we can say that as the common ion effect increases the solubility will also increases. Hence, the reason is also false. Hence, both A and R are false statement.
SRMJEEE - 2014
Ionic Equilibrium
230038
The pH of the resultant solution obtained by mixing 20 mL of 0.01 M HCl and 20 mL of \(0.005 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2\) is
1 2
2 0
3 1
4 7
5 5
Explanation:
Total no. of moles present in 20 ml of 0.01 M HCl . \[ \begin{aligned} & =20 \times 0.01 \\ & =20 \times 1 \times 10^{-2} \\ & =2 \times 10^{-1} \\ & =0.2 \mathrm{moles} \end{aligned} \] Similarly, Total no. of moles present in 20 ml of 0.005 M \[ \begin{aligned} \mathrm{Ca}(\mathrm{OH})_2 & =20 \times 0.005 \times 2 \\ & =40 \times 0.005 \\ & =4 \times 5 \times 10^{-2} \\ & =20 \times 10^{-2} \\ & =0.2 \text { moles. } \end{aligned} \] So, pH of the resultant solution will be 7 .
Kerala-CEE-2020
Ionic Equilibrium
230041
The ionization constant of ammonia hydroxide is \(1.77 \times 10^{-5}\) at 298 K . Hydrolysis constant of ammonium chloride is
1 \(6.50 \times 10^{-12}\)
2 \(5.65 \times 10^{-13}\)
3 \(5.65 \times 10^{-12}\)
4 \(5.65 \times 10^{-10}\)
Explanation:
\(\mathrm{NH}_4 \mathrm{Cl}\) is a salt of strong acid and weak Base, so hydrolysis constant is \[ \begin{aligned} & \quad \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_w}{\mathrm{~K}_{\mathrm{b}}} \\ & \text { Given }-\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5} \\ & \mathrm{~K}_{\mathrm{W}}=10^{-14} \\ & \therefore \quad \mathrm{~K}_{\mathrm{h}}=\frac{10^{-14}}{1.77 \times 10^{-5}}=0.565 \times 10^{-9} \\ & \text { or } \quad \mathrm{K}_{\mathrm{h}}=5.65 \times 10^{-10} \end{aligned} \]
NEET-2009
Ionic Equilibrium
230042
What is the \([\mathrm{OH}]\) in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 of \(0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2\) ?
1 0.40 M
2 0.0050 M
3 0.12 M
4 0.10 M
Explanation:
Milimoles of \(\mathrm{H}^{+}\)produced \(=20 \times 0.05=1\) milimoles of \(\mathrm{OH}^{-}\)produced \(=30 \times 0.1 \times 2=6\) \(\because \mathrm{Ba}(\mathrm{OH})_2 \square \quad \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}\) \(\therefore \quad\) Milimoles of \(\mathrm{OH}^{-}\)remaining in solution \(6-1=5\) Total volume of solution \(=20+30=50 \mathrm{~mL}\) \(\therefore \quad[\mathrm{OH}]=\frac{5}{50}=0.1 \mathrm{M}\)
230034
When common salt is added to a saturated solution of soap, soap is precipitated. This is based on the principle of
1 common ion effect
2 solubility product
3 adsorption form solution
4 peptisation.
Explanation:
The saturated solution of soap (RCOONa), show precipitation addition of NaCl , because an increase in \(\mathrm{Na}^{+}\)concentration helps in crossing over ionic product to their \(\mathrm{K}_{\mathrm{sp}}\) values. \[ \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}(\mathrm{~s}) \square \quad \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq}) \] When NaCl is added, \(\left[\mathrm{Na}^{+}\right]\)increases due to which ionic product, \(\left[\mathrm{RCOO}^{-}\right]\left[\mathrm{Na}^{+}\right]\)also increases which results in precipitation of soap. It can also be explained on the basis of common ion effect. Among the given compound the \(\mathrm{NH}_3\) is the most basic. Hence has highest proton affinity.
COMEDK-2018
Ionic Equilibrium
230035
Assertion (A) : \(\mathrm{Sb}_2 \mathrm{~S}_3\) is not soluble in yellow ammonium sulphide. Reason (R) : The common ion effect due to \(\mathrm{S}^{2-}\) ions reduces the stability of \(\mathbf{S b}_2 \mathrm{~S}_3\).
1 Both A and R are true and R is the correct explanation of the A .
2 Both A and R are true but R is not the correct explanation of the \(A\).
3 A is true, but R is false.
4 Both A and R are false statements.
Explanation:
(A) \(\mathrm{Sb}_2 \mathrm{~S}_3\) is soluble in yellow ammonium sulphide. \[ \mathrm{Sb}_2 \mathrm{~S}_3+3\left(\mathrm{NH}_4\right)_2 \mathrm{~S} \longrightarrow 2\left(\mathrm{NH}_4\right)_3 \mathrm{SbS}_3 \] Hence, the statement is false. (R) This is due to the fact that when the colour changes from white to yellow the number of sulphur increases. therefore the ions of the sulphur atoms increases i.e. \(\mathrm{S}^{2-}\) ions. So, as the amount of the sulphur ions increases the common ion effect also increases. Therefore, we can say that as the common ion effect increases the solubility will also increases. Hence, the reason is also false. Hence, both A and R are false statement.
SRMJEEE - 2014
Ionic Equilibrium
230038
The pH of the resultant solution obtained by mixing 20 mL of 0.01 M HCl and 20 mL of \(0.005 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2\) is
1 2
2 0
3 1
4 7
5 5
Explanation:
Total no. of moles present in 20 ml of 0.01 M HCl . \[ \begin{aligned} & =20 \times 0.01 \\ & =20 \times 1 \times 10^{-2} \\ & =2 \times 10^{-1} \\ & =0.2 \mathrm{moles} \end{aligned} \] Similarly, Total no. of moles present in 20 ml of 0.005 M \[ \begin{aligned} \mathrm{Ca}(\mathrm{OH})_2 & =20 \times 0.005 \times 2 \\ & =40 \times 0.005 \\ & =4 \times 5 \times 10^{-2} \\ & =20 \times 10^{-2} \\ & =0.2 \text { moles. } \end{aligned} \] So, pH of the resultant solution will be 7 .
Kerala-CEE-2020
Ionic Equilibrium
230041
The ionization constant of ammonia hydroxide is \(1.77 \times 10^{-5}\) at 298 K . Hydrolysis constant of ammonium chloride is
1 \(6.50 \times 10^{-12}\)
2 \(5.65 \times 10^{-13}\)
3 \(5.65 \times 10^{-12}\)
4 \(5.65 \times 10^{-10}\)
Explanation:
\(\mathrm{NH}_4 \mathrm{Cl}\) is a salt of strong acid and weak Base, so hydrolysis constant is \[ \begin{aligned} & \quad \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_w}{\mathrm{~K}_{\mathrm{b}}} \\ & \text { Given }-\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5} \\ & \mathrm{~K}_{\mathrm{W}}=10^{-14} \\ & \therefore \quad \mathrm{~K}_{\mathrm{h}}=\frac{10^{-14}}{1.77 \times 10^{-5}}=0.565 \times 10^{-9} \\ & \text { or } \quad \mathrm{K}_{\mathrm{h}}=5.65 \times 10^{-10} \end{aligned} \]
NEET-2009
Ionic Equilibrium
230042
What is the \([\mathrm{OH}]\) in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 of \(0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2\) ?
1 0.40 M
2 0.0050 M
3 0.12 M
4 0.10 M
Explanation:
Milimoles of \(\mathrm{H}^{+}\)produced \(=20 \times 0.05=1\) milimoles of \(\mathrm{OH}^{-}\)produced \(=30 \times 0.1 \times 2=6\) \(\because \mathrm{Ba}(\mathrm{OH})_2 \square \quad \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}\) \(\therefore \quad\) Milimoles of \(\mathrm{OH}^{-}\)remaining in solution \(6-1=5\) Total volume of solution \(=20+30=50 \mathrm{~mL}\) \(\therefore \quad[\mathrm{OH}]=\frac{5}{50}=0.1 \mathrm{M}\)
230034
When common salt is added to a saturated solution of soap, soap is precipitated. This is based on the principle of
1 common ion effect
2 solubility product
3 adsorption form solution
4 peptisation.
Explanation:
The saturated solution of soap (RCOONa), show precipitation addition of NaCl , because an increase in \(\mathrm{Na}^{+}\)concentration helps in crossing over ionic product to their \(\mathrm{K}_{\mathrm{sp}}\) values. \[ \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}(\mathrm{~s}) \square \quad \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq}) \] When NaCl is added, \(\left[\mathrm{Na}^{+}\right]\)increases due to which ionic product, \(\left[\mathrm{RCOO}^{-}\right]\left[\mathrm{Na}^{+}\right]\)also increases which results in precipitation of soap. It can also be explained on the basis of common ion effect. Among the given compound the \(\mathrm{NH}_3\) is the most basic. Hence has highest proton affinity.
COMEDK-2018
Ionic Equilibrium
230035
Assertion (A) : \(\mathrm{Sb}_2 \mathrm{~S}_3\) is not soluble in yellow ammonium sulphide. Reason (R) : The common ion effect due to \(\mathrm{S}^{2-}\) ions reduces the stability of \(\mathbf{S b}_2 \mathrm{~S}_3\).
1 Both A and R are true and R is the correct explanation of the A .
2 Both A and R are true but R is not the correct explanation of the \(A\).
3 A is true, but R is false.
4 Both A and R are false statements.
Explanation:
(A) \(\mathrm{Sb}_2 \mathrm{~S}_3\) is soluble in yellow ammonium sulphide. \[ \mathrm{Sb}_2 \mathrm{~S}_3+3\left(\mathrm{NH}_4\right)_2 \mathrm{~S} \longrightarrow 2\left(\mathrm{NH}_4\right)_3 \mathrm{SbS}_3 \] Hence, the statement is false. (R) This is due to the fact that when the colour changes from white to yellow the number of sulphur increases. therefore the ions of the sulphur atoms increases i.e. \(\mathrm{S}^{2-}\) ions. So, as the amount of the sulphur ions increases the common ion effect also increases. Therefore, we can say that as the common ion effect increases the solubility will also increases. Hence, the reason is also false. Hence, both A and R are false statement.
SRMJEEE - 2014
Ionic Equilibrium
230038
The pH of the resultant solution obtained by mixing 20 mL of 0.01 M HCl and 20 mL of \(0.005 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2\) is
1 2
2 0
3 1
4 7
5 5
Explanation:
Total no. of moles present in 20 ml of 0.01 M HCl . \[ \begin{aligned} & =20 \times 0.01 \\ & =20 \times 1 \times 10^{-2} \\ & =2 \times 10^{-1} \\ & =0.2 \mathrm{moles} \end{aligned} \] Similarly, Total no. of moles present in 20 ml of 0.005 M \[ \begin{aligned} \mathrm{Ca}(\mathrm{OH})_2 & =20 \times 0.005 \times 2 \\ & =40 \times 0.005 \\ & =4 \times 5 \times 10^{-2} \\ & =20 \times 10^{-2} \\ & =0.2 \text { moles. } \end{aligned} \] So, pH of the resultant solution will be 7 .
Kerala-CEE-2020
Ionic Equilibrium
230041
The ionization constant of ammonia hydroxide is \(1.77 \times 10^{-5}\) at 298 K . Hydrolysis constant of ammonium chloride is
1 \(6.50 \times 10^{-12}\)
2 \(5.65 \times 10^{-13}\)
3 \(5.65 \times 10^{-12}\)
4 \(5.65 \times 10^{-10}\)
Explanation:
\(\mathrm{NH}_4 \mathrm{Cl}\) is a salt of strong acid and weak Base, so hydrolysis constant is \[ \begin{aligned} & \quad \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_w}{\mathrm{~K}_{\mathrm{b}}} \\ & \text { Given }-\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5} \\ & \mathrm{~K}_{\mathrm{W}}=10^{-14} \\ & \therefore \quad \mathrm{~K}_{\mathrm{h}}=\frac{10^{-14}}{1.77 \times 10^{-5}}=0.565 \times 10^{-9} \\ & \text { or } \quad \mathrm{K}_{\mathrm{h}}=5.65 \times 10^{-10} \end{aligned} \]
NEET-2009
Ionic Equilibrium
230042
What is the \([\mathrm{OH}]\) in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 of \(0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2\) ?
1 0.40 M
2 0.0050 M
3 0.12 M
4 0.10 M
Explanation:
Milimoles of \(\mathrm{H}^{+}\)produced \(=20 \times 0.05=1\) milimoles of \(\mathrm{OH}^{-}\)produced \(=30 \times 0.1 \times 2=6\) \(\because \mathrm{Ba}(\mathrm{OH})_2 \square \quad \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}\) \(\therefore \quad\) Milimoles of \(\mathrm{OH}^{-}\)remaining in solution \(6-1=5\) Total volume of solution \(=20+30=50 \mathrm{~mL}\) \(\therefore \quad[\mathrm{OH}]=\frac{5}{50}=0.1 \mathrm{M}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ionic Equilibrium
230034
When common salt is added to a saturated solution of soap, soap is precipitated. This is based on the principle of
1 common ion effect
2 solubility product
3 adsorption form solution
4 peptisation.
Explanation:
The saturated solution of soap (RCOONa), show precipitation addition of NaCl , because an increase in \(\mathrm{Na}^{+}\)concentration helps in crossing over ionic product to their \(\mathrm{K}_{\mathrm{sp}}\) values. \[ \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}(\mathrm{~s}) \square \quad \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq}) \] When NaCl is added, \(\left[\mathrm{Na}^{+}\right]\)increases due to which ionic product, \(\left[\mathrm{RCOO}^{-}\right]\left[\mathrm{Na}^{+}\right]\)also increases which results in precipitation of soap. It can also be explained on the basis of common ion effect. Among the given compound the \(\mathrm{NH}_3\) is the most basic. Hence has highest proton affinity.
COMEDK-2018
Ionic Equilibrium
230035
Assertion (A) : \(\mathrm{Sb}_2 \mathrm{~S}_3\) is not soluble in yellow ammonium sulphide. Reason (R) : The common ion effect due to \(\mathrm{S}^{2-}\) ions reduces the stability of \(\mathbf{S b}_2 \mathrm{~S}_3\).
1 Both A and R are true and R is the correct explanation of the A .
2 Both A and R are true but R is not the correct explanation of the \(A\).
3 A is true, but R is false.
4 Both A and R are false statements.
Explanation:
(A) \(\mathrm{Sb}_2 \mathrm{~S}_3\) is soluble in yellow ammonium sulphide. \[ \mathrm{Sb}_2 \mathrm{~S}_3+3\left(\mathrm{NH}_4\right)_2 \mathrm{~S} \longrightarrow 2\left(\mathrm{NH}_4\right)_3 \mathrm{SbS}_3 \] Hence, the statement is false. (R) This is due to the fact that when the colour changes from white to yellow the number of sulphur increases. therefore the ions of the sulphur atoms increases i.e. \(\mathrm{S}^{2-}\) ions. So, as the amount of the sulphur ions increases the common ion effect also increases. Therefore, we can say that as the common ion effect increases the solubility will also increases. Hence, the reason is also false. Hence, both A and R are false statement.
SRMJEEE - 2014
Ionic Equilibrium
230038
The pH of the resultant solution obtained by mixing 20 mL of 0.01 M HCl and 20 mL of \(0.005 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2\) is
1 2
2 0
3 1
4 7
5 5
Explanation:
Total no. of moles present in 20 ml of 0.01 M HCl . \[ \begin{aligned} & =20 \times 0.01 \\ & =20 \times 1 \times 10^{-2} \\ & =2 \times 10^{-1} \\ & =0.2 \mathrm{moles} \end{aligned} \] Similarly, Total no. of moles present in 20 ml of 0.005 M \[ \begin{aligned} \mathrm{Ca}(\mathrm{OH})_2 & =20 \times 0.005 \times 2 \\ & =40 \times 0.005 \\ & =4 \times 5 \times 10^{-2} \\ & =20 \times 10^{-2} \\ & =0.2 \text { moles. } \end{aligned} \] So, pH of the resultant solution will be 7 .
Kerala-CEE-2020
Ionic Equilibrium
230041
The ionization constant of ammonia hydroxide is \(1.77 \times 10^{-5}\) at 298 K . Hydrolysis constant of ammonium chloride is
1 \(6.50 \times 10^{-12}\)
2 \(5.65 \times 10^{-13}\)
3 \(5.65 \times 10^{-12}\)
4 \(5.65 \times 10^{-10}\)
Explanation:
\(\mathrm{NH}_4 \mathrm{Cl}\) is a salt of strong acid and weak Base, so hydrolysis constant is \[ \begin{aligned} & \quad \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_w}{\mathrm{~K}_{\mathrm{b}}} \\ & \text { Given }-\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5} \\ & \mathrm{~K}_{\mathrm{W}}=10^{-14} \\ & \therefore \quad \mathrm{~K}_{\mathrm{h}}=\frac{10^{-14}}{1.77 \times 10^{-5}}=0.565 \times 10^{-9} \\ & \text { or } \quad \mathrm{K}_{\mathrm{h}}=5.65 \times 10^{-10} \end{aligned} \]
NEET-2009
Ionic Equilibrium
230042
What is the \([\mathrm{OH}]\) in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 of \(0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2\) ?
1 0.40 M
2 0.0050 M
3 0.12 M
4 0.10 M
Explanation:
Milimoles of \(\mathrm{H}^{+}\)produced \(=20 \times 0.05=1\) milimoles of \(\mathrm{OH}^{-}\)produced \(=30 \times 0.1 \times 2=6\) \(\because \mathrm{Ba}(\mathrm{OH})_2 \square \quad \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}\) \(\therefore \quad\) Milimoles of \(\mathrm{OH}^{-}\)remaining in solution \(6-1=5\) Total volume of solution \(=20+30=50 \mathrm{~mL}\) \(\therefore \quad[\mathrm{OH}]=\frac{5}{50}=0.1 \mathrm{M}\)
230034
When common salt is added to a saturated solution of soap, soap is precipitated. This is based on the principle of
1 common ion effect
2 solubility product
3 adsorption form solution
4 peptisation.
Explanation:
The saturated solution of soap (RCOONa), show precipitation addition of NaCl , because an increase in \(\mathrm{Na}^{+}\)concentration helps in crossing over ionic product to their \(\mathrm{K}_{\mathrm{sp}}\) values. \[ \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COONa}(\mathrm{~s}) \square \quad \mathrm{C}_{17} \mathrm{H}_{35} \mathrm{COO}^{-}(\mathrm{aq})+\mathrm{Na}^{+}(\mathrm{aq}) \] When NaCl is added, \(\left[\mathrm{Na}^{+}\right]\)increases due to which ionic product, \(\left[\mathrm{RCOO}^{-}\right]\left[\mathrm{Na}^{+}\right]\)also increases which results in precipitation of soap. It can also be explained on the basis of common ion effect. Among the given compound the \(\mathrm{NH}_3\) is the most basic. Hence has highest proton affinity.
COMEDK-2018
Ionic Equilibrium
230035
Assertion (A) : \(\mathrm{Sb}_2 \mathrm{~S}_3\) is not soluble in yellow ammonium sulphide. Reason (R) : The common ion effect due to \(\mathrm{S}^{2-}\) ions reduces the stability of \(\mathbf{S b}_2 \mathrm{~S}_3\).
1 Both A and R are true and R is the correct explanation of the A .
2 Both A and R are true but R is not the correct explanation of the \(A\).
3 A is true, but R is false.
4 Both A and R are false statements.
Explanation:
(A) \(\mathrm{Sb}_2 \mathrm{~S}_3\) is soluble in yellow ammonium sulphide. \[ \mathrm{Sb}_2 \mathrm{~S}_3+3\left(\mathrm{NH}_4\right)_2 \mathrm{~S} \longrightarrow 2\left(\mathrm{NH}_4\right)_3 \mathrm{SbS}_3 \] Hence, the statement is false. (R) This is due to the fact that when the colour changes from white to yellow the number of sulphur increases. therefore the ions of the sulphur atoms increases i.e. \(\mathrm{S}^{2-}\) ions. So, as the amount of the sulphur ions increases the common ion effect also increases. Therefore, we can say that as the common ion effect increases the solubility will also increases. Hence, the reason is also false. Hence, both A and R are false statement.
SRMJEEE - 2014
Ionic Equilibrium
230038
The pH of the resultant solution obtained by mixing 20 mL of 0.01 M HCl and 20 mL of \(0.005 \mathrm{M} \mathrm{Ca}(\mathrm{OH})_2\) is
1 2
2 0
3 1
4 7
5 5
Explanation:
Total no. of moles present in 20 ml of 0.01 M HCl . \[ \begin{aligned} & =20 \times 0.01 \\ & =20 \times 1 \times 10^{-2} \\ & =2 \times 10^{-1} \\ & =0.2 \mathrm{moles} \end{aligned} \] Similarly, Total no. of moles present in 20 ml of 0.005 M \[ \begin{aligned} \mathrm{Ca}(\mathrm{OH})_2 & =20 \times 0.005 \times 2 \\ & =40 \times 0.005 \\ & =4 \times 5 \times 10^{-2} \\ & =20 \times 10^{-2} \\ & =0.2 \text { moles. } \end{aligned} \] So, pH of the resultant solution will be 7 .
Kerala-CEE-2020
Ionic Equilibrium
230041
The ionization constant of ammonia hydroxide is \(1.77 \times 10^{-5}\) at 298 K . Hydrolysis constant of ammonium chloride is
1 \(6.50 \times 10^{-12}\)
2 \(5.65 \times 10^{-13}\)
3 \(5.65 \times 10^{-12}\)
4 \(5.65 \times 10^{-10}\)
Explanation:
\(\mathrm{NH}_4 \mathrm{Cl}\) is a salt of strong acid and weak Base, so hydrolysis constant is \[ \begin{aligned} & \quad \mathrm{K}_{\mathrm{h}}=\frac{\mathrm{K}_w}{\mathrm{~K}_{\mathrm{b}}} \\ & \text { Given }-\mathrm{K}_{\mathrm{b}}\left(\mathrm{NH}_4 \mathrm{OH}\right)=1.77 \times 10^{-5} \\ & \mathrm{~K}_{\mathrm{W}}=10^{-14} \\ & \therefore \quad \mathrm{~K}_{\mathrm{h}}=\frac{10^{-14}}{1.77 \times 10^{-5}}=0.565 \times 10^{-9} \\ & \text { or } \quad \mathrm{K}_{\mathrm{h}}=5.65 \times 10^{-10} \end{aligned} \]
NEET-2009
Ionic Equilibrium
230042
What is the \([\mathrm{OH}]\) in the final solution prepared by mixing 20.0 mL of 0.050 M HCl with 30.0 of \(0.10 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_2\) ?
1 0.40 M
2 0.0050 M
3 0.12 M
4 0.10 M
Explanation:
Milimoles of \(\mathrm{H}^{+}\)produced \(=20 \times 0.05=1\) milimoles of \(\mathrm{OH}^{-}\)produced \(=30 \times 0.1 \times 2=6\) \(\because \mathrm{Ba}(\mathrm{OH})_2 \square \quad \mathrm{Ba}^{2+}+2 \mathrm{OH}^{-}\) \(\therefore \quad\) Milimoles of \(\mathrm{OH}^{-}\)remaining in solution \(6-1=5\) Total volume of solution \(=20+30=50 \mathrm{~mL}\) \(\therefore \quad[\mathrm{OH}]=\frac{5}{50}=0.1 \mathrm{M}\)