Home
Class 11
CHEMISTRY
For a general reaction given below, the ...

For a general reaction given below, the value of solubility product can be given us
`{:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):}`
`K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y)`
Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation `[H^(+)]` ion, `[OH^(-)`] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals
Potussium chromate is slowly aded toa solution containing 0.20M Ag `NO_(3)`, and 0.20M `Ba(NO_(3))_(2)`. Describe what happensif the `K_(sp)` for `Ag_(2),CrO_(4)`, is `1.1 xx 10^(-12)` and the `K_(sp)` of `BaCiO_(4)`, is `1.2 xx 10^(-10)`,

A

The ` Ag_2CrO_4` pecipitates first out of solution and then `BaCrO_4 ` preciptates.

B

The `BaCrO_4` pecipitates first out of solution and then ` Ag_2CrO_4` preciptates

C

Both `Ag_2CrO_4 and BaCrO_4` precipitate simultaneously out of solution

D

Neither `Ag_2CrO_4 " nor " BaCrO_4` precipitates

Text Solution

Verified by Experts

The correct Answer is:
A
` [CrO_4^(2-)]_(Ag^(+)) =(Ksp)/( [Ag^(+) ]^(2)) =(1.1xx 10^(-12))/(0.2) =5.5 xx 10 ^(-12) `
` [CrO_4^(2-) ]_(Ba^(+2) ) =(Ksp)/([Ba^(+2)]) =(1.2xx 10^(-10))/( 0.2 0 ) = 6 xx 10^(-10) `
` therefore Ag_2 Cr O_4 ` ppts first
Promotional Banner

Topper's Solved these Questions

  • IONIC EQUILIBRIUM

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL -II PRACTICE SHEET (ADVANCED) (Integer Type Questions))|8 Videos

  • IONIC EQUILIBRIUM

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE (LEVEL -II PRACTICE SHEET (ADVANCED) (More than One correct answer Type Questions))|4 Videos

  • HYDROGEN AND ITS COMPOUNDS

    AAKASH SERIES|Exercise QUESTIONS FOR DESCRIPTIVE ANSWERS|20 Videos

  • ISOMERISM

    AAKASH SERIES|Exercise ADDITIONAL PRACTICE EXERCISE ( PRACTICE SHEET (ADVANCED) INTEGER TYPE QUESTIONS)|10 Videos

Similar Questions

Explore conceptually related problems

For a general reaction given below, the value of solubility product can be given us {:(A_(x)B_(y),=xA^(+y),+yB^(-x)),(a,0,0),(a-s,xs,ys):} K_(sp)=(xs)^(x).(ys)^(y) (or) K_(sp)=x^(x)y^(y) (S)^(x+y) Solubility product gives us not only an idea about the solubility of an electrolyte in a solvent but also helps in explaining concept of precipitation and calculation [H^(+)] ion, [OH^(-) ] ion. It is also useful in qualitative analysis for the idetification and separation of basic radicals What is the molar solubility of Cu(OH)_(2) , in 1.0 M NH_(3) if the deep blue complex ion [Cu(NH_(3))_(4)]^(2+) is formed. The K_(sp), of Cu(OH)_(2) , is 1.6xx 10^(-19) and K_(3) , of [Cu(NH_(3))_(4) is 1.1 xx 10^(13)

Give any two application of solubility product and common ion effect in qualitative analysis.

Let the solubility of an aqueous solution of Mg(OH)_(2) , be " X^(@) then its K_(sp) is

Using information given in the figure, calculate the value of x and y.

K_(sp) of M(OH)_(x) , is 27 xx 10^(-12) and its solubility in water is 10^(-3) mol litre^(-1) . Find the value of X

Using information given in the figure, calculate the value of x and y

Two line are given (x-2y)^2+k(x-2y)=0 . The value of k, so that the distance between then is 3, is

A solution which remains in equilibrium with undissolved solute is said to be saturated. The concentration of a saturated solution at a given temperature is called solubility. The product of concentration of ions in a saturated solution of an electrolyte at a given temperature, is called solubility product (K_(sp)) . For the electrolyte, A_(x),B_(y),:A_(x),B_(y(s)) rarr xA^(y+)+ y^(Bx-) , with solubility S, the solubility product (K_(sp)) =x^(x)xxy^(y) xx s^(x+y) . While calculating the solubility of a sparingly soluble salt in the presence of some strong electrolyte containing a common ion, the common ion concentration is practically equal to that of strong electrolyte. If in a solution, the ionic product of an clectrolyte exceeds its K_(sp) , value at a particular temperature, then precipitation occurs. The solubility of BaSO_(4) , in 0.1 M BaCl_(2) , solution is (K_(sp) , of BaSO_(4), = 1.5 xx 10^(-9))