Physical and chemical equilibrium can respond to a change in their pressure, temperature, and concentration of reactants and products. To describe the change in the equilibrium we have a principle named Le Chatelier principle. According to this principle, even if we make some changes in equilibrium, then also the system even re-establishes the equilibrium by undoing the effect. If we add `SO_(4)^(2-)` ion to a saturated solution of `Ag_(2)SO_(4)`, it will result in a//anA. Result in an increase in `Ag^(o+)` concentrationB. Result in a decrease in `Ag^(o+)` concentrationC. Shift `Ag^(o+)` ions from solid `Ag_(2)CrO_(4)` into solution.D. Result in a decrease the `CrO_(4)^(2-)` ion concentration in the solution.

Physical and chemical equilibrium can respond to a change in their pre

1.	Physical and chemical equilibrium can respond to a change in their pressure, temperature, and concentration of reactants and products. To describe the change in the equilibrium we have a principle named Le Chatelier principle. According to this principle, even if we make some changes in equilibrium, then also the system even re-establishes the equilibrium by undoing the effect. If we add `SO_(4)^(2-)` ion to a saturated solution of `Ag_(2)SO_(4)`, it will result in a//anA. Result in an increase in `Ag^(o+)` concentrationB. Result in a decrease in `Ag^(o+)` concentrationC. Shift `Ag^(o+)` ions from solid `Ag_(2)CrO_(4)` into solution.D. Result in a decrease the `CrO_(4)^(2-)` ion concentration in the solution.
Answer» Correct Answer - B Due to common ion `(CrO_(4)^(2-))`, the ionisation of `Ag_(2)CrO_(4)` is supressed and solubility of `Ag_(2)CrO_(4)` decreases i.e., `[Ag^(o+)]` decreases.

Discussion

No Comment Found

Related InterviewSolutions

100ml of 0.1M `H_(3)PO_(4)` is titrated with 0.05 M NaOH solution till 2nd equivalent point. Then, resultant solution was mixed with 10 ml of 0.5 MHCI solution `K_(1)=10^(-3)M` `K_(2)=10^(-8)M` `K_(3)=10^(-13)M` pH at 2nd equivalent point will be :A. `5.5`B. `10.5`C. `8`D. `7`
The conjugate bas of hydrazoic acid isA. `N_(2)H_(4)`B. `N_(2)H_(5^(+))`C. `N_(3)^(-)`D. `NH_(2)OH`
Calcualte the percentage hydrolysis of `10^(-3)M N_(2)^(o+)H_(5)C1^(Theta)` (hydrazinium chloride), salt contining acid ion conjugate to hydrazine base `(NH_(2)NH_(2)). K_(b)` for `N_(2)H_(4) = 1.0 xx 10^(-6)`.
Calculate the hydrolysis constant `(K_(h))` and degree of hydrloysis `(h)` of `NH_(4)C1` in `0.1M` solution. `K_(b) = 2.0 xx 10^(-5)`. Calculate the `[overset(Theta)OH]` ions in the solution.
A trpical aspirin tablet constains 324 mg of aspirin (acetyl salicylic acid `C_(9)H_(8)O_(4)`) a monoprotic acid having `K_(a)=3.0xx10^(-4)` . What is the degree of dissociation of salt and pH of the solution, if two aspirin tables are dissolved to prepare 300 mL solution in water ?
What is the entropy that takes place in conversion of 1 mol of `alpha-`tin to 1 mol `beta -` tin at `17^(@)C`, 1 atm pressure if enthalpy of transition is 2.90 kJ `mol^(-1)` ?A. `1.0 kJ mol^(-1)`B. `7.32 JK^(-1) mol^(-1)`C. `10JK^(-1)mol^(-1)`D. `100JK mol^(-1)`
We represent various reagents pictorially such that, The relative area of the squars represents the relative volume of the reagents used and the overlapping region represents the relative amount of mixing of the reagents. (Use log 5=0.7,log2=0.3,log3=0.5,`d_(H_2O)=1gml^(-1)`) For a solution represented by Which is/are correct?A. Ph=1B. `alpha` of weak acid =`5xx10^(-4)`C. `alpha` of `H_(2)O=9xx10^(-15)`D. Ph=1.7
Determine the concentration of `NH_(3)` solution whose one litre can dissolve `0.10` mole AgCI. `K_(SP)` of AgCI and `K_(f)` of `Ag(NH_(3))_(2)^(+)` are `1.0xx10^(-10)M^(2)` and `1.6xx10^(7)M^(-2)` respectively.
If `500 ml` of `0.4 M AgNO_(3)` is mixed with `50 ml` of `2 M NH_(3)` solution then what is the concentration of `[Ag(NH_(3))]^(+)` in solution `(K_(t), [Ag_(NH_(3))]^(+) = 10^(3), K_(f_(2)) [Ag(NH_(3))_(2)]^(+)= 10^(4))`A. `3.33 xx 10^(-7) M`B. `3.33 xx 10^(-5) M`C. `3 xx 10^(-4) M`D. `10^(-7) M`
a. Calculate `[Ag^(o+)]` in a solution of `[Ag^(o+)]` in a solution of `[Ag(NH_(3))_(2)^(o+)]` prepared by adding `1.0 xx 10^(-3)mol AgNO_(3)` to `1.0L of 1.0M NH_(3)` solution `K_(f) Ag(NH_(3))_(2)^(o+) = 10^(8)`. b. Calculate `[Ag^(o+)]` which is in equilibrium with `0.15M [Ag(NH_(3))_(2)]^(o+)` and `1.5 NH_(3)`.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment