The `K_(SP)` of `Ag_(2)C_(2)O_(4)` at `25^(@)C` is `1.29xx10^(-11)mol^(3)L^(-3)`. A solution of `K_(2)C_(2)O_(4)` containing `0.152` mole in 500 mL water is shaken at `25^(@)C` with excess of `Ag_(2)CO_(3)` till the equilbirium is reached. `Ag_(2)CO_(3)+K_(2)C_(2)O_(4)hArrAg_(2)C_(2)O_(4)+K_(2)CO_(3)` Ar equilibrium the solution contains `0.0358` mole of `K_(2)CO_(3)`. Assuming degree of dissociation of `K_(2)C_(2)O_(4)` and `K_(2)CO_(3)` to be same, calculate `K_(SP)` of `Ag_(2)CO_(3)`.

The `K_(SP)` of `Ag_(2)C_(2)O_(4)` at `25^(@)C` is `1.29xx10^(-11)mol^

1.	The `K_(SP)` of `Ag_(2)C_(2)O_(4)` at `25^(@)C` is `1.29xx10^(-11)mol^(3)L^(-3)`. A solution of `K_(2)C_(2)O_(4)` containing `0.152` mole in 500 mL water is shaken at `25^(@)C` with excess of `Ag_(2)CO_(3)` till the equilbirium is reached. `Ag_(2)CO_(3)+K_(2)C_(2)O_(4)hArrAg_(2)C_(2)O_(4)+K_(2)CO_(3)` Ar equilibrium the solution contains `0.0358` mole of `K_(2)CO_(3)`. Assuming degree of dissociation of `K_(2)C_(2)O_(4)` and `K_(2)CO_(3)` to be same, calculate `K_(SP)` of `Ag_(2)CO_(3)`.
Answer» Correct Answer - `9.67x10^(-11)`

Discussion

No Comment Found

Related InterviewSolutions

`1`mol of `Cl_(2)` and `3` mol of `PCl_(5)` are placed in a `100 L` vessel heated to `227^(@)C`. The equilibrium pressure is `2.05` atm. Assuming ideal behaviour, calculate the degree of dissociation for `PCl_(5)` and `K_(p)` for the reaction. `PCl_(5)(g) hArr PCl_(3)(g)+Cl_(2)(g)`
The change in pressure will not affect the equilibrium constant forA. `N_(2)+3H_(2) hArr 2NH_(3)`B. `PCI_(5) hArr PCI_(3) +CI_(2)`C. `H_(2)+I_(2) hArr 2HI`D. All of these
The equation for the reaction in the figure given below is `AB_(5)(g)+Heat hArrAB_(3)(g)+B_(2)(g)` At time three minutes what change was imposed into the equilibrium?A. Pressure was increasedB. Temperature was increasedC. `B_(2)` was added to the systemD. `AB_(3)` was added to the system
In a 500mL falsk, the degree of dissociation of `PCI_(5)` at equilibrium is `40%` and the initial amount is 5 moles. The value of equilibrium constant in mol `L^(-1)` for the decomposition of `PCI_(5)` isA. `2.33`B. `2.66`C. `5.32`D. `4.66`
Consider the following reversible system: `A(g)+2B(g)hArrC(g)+2D(g)` At equilibrium there are 1.0 mole of A and 2.0 moles of each B, C and D present. If 2.0 moles of B is added so that moles of A and D do not change?
To determine molar solubility of an unknown metal hydroxide, `M(OH)_(3)` in `0.00101M NaOH,` Jane adds `0.250g` of `M(OH)_(3)` solid to `200mL` of a `0.00101M NaOH` solution and stirred for a long time.A cloudy solution is formed. The undissolved hydroxide is separated from the solution by filtratin `25.00mL` of filtrate is pipetted into a flask `30mL` of water is added adn then the solution is titrated with `0.00444M HCl` solution `13mL` of `HCl` is used for this titration. If Jane carelessly added `20mL` of water tothe flask instead of `30mL` how would this affect the determination of solubility product `(K_(sp))` and molar solubility? (1) The value of `K_(sp)` determined by Jane would be same as true value (2) The value of `K_(sp)` determined by Jane would be higher than the true value. (3) The value of `K_(sp)` determined by Jane would be lower as true value. (4) The molar solubility of `M(OH)_(3)` determined by Jane woud be same as true value. (5) The molar solubility of `M(OH)_(3)` determined by Jane would be higher than true value. (6) The molar solubililty of `M(OH)_(3)` determined by Jane would be lower than the true valueA. 1 and 4B. 2 and 5C. 3 and 6D. 1 and 6
An aqueous solution of a metal bromide `MBr_(2)(0.05M)` is saturated with `H_(2)S`. What is the minimum pH at which MS will precipitate ? `K_(SP)` for `M S= 6.0xx10^(-21)` . Concentration of saturqated `H_(2)S=0.1M, K_(1)=10^(-7)and K_(2)=1.3xx10^(-13)` for `H_(2)S` .
`(CH_(3)(OH)_(2)^(1))` is..acidic than `(CH_(3)NH_(3)^(+))`.
The conjugate acid of amide ion `(NH_(2)^(-))` isA. `NH_(3)`B. `NH_(2)OH`C. `NH_(4)^(+)`D. `N_(2)H_(4)`
Which of the following relations represent correct relation between standard electrode potential and equilibrium constant ? I. `logK = (nFE^(@))/(2303 RT)` II. `K = e^(-(nFE^(@))/(RT))` III. `log K = (-nFE^(@))/(2303 RT)` IV. `log K = 0.4342 (nFE^(@))/(RT)`A. I,II and IIIB. II and IIIC. I, II and IVD. I and IV

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment