Question

Concentration of Ag^+ ions in saturated solution of Ag2CrO4 at 20°C is 1.5×10^-4 mol/ L At 20°C the solubility product of Ag2CrO4 is

Answer 1

Answer:- solubility product is $1.69*10^-^1^2$ .

Solution:- $Ag_2CrO_4$ gives the ions as shown in the below equation:

$Ag_2CrO_4(s)\rightarrow 2Ag^+(aq)+CrO_4^2^-(aq)$

If the molar solubility of $Ag_2CrO_4$ is s then $Ag^+$ is 2s and $CrO_4^2^-$ is s. (It's because there is 2:1 ration between ions)

Silver ion concentration in saturated solution is given as $1.5*10^-^4\frac{mol}{L}$ . It means, $2s=1.5*10^-^4\frac{mol}{L}$ .

$s=7.5*10^-^5\frac{mol}{L}$

So, $CrO_4^2^-=7.5*10^-^5\frac{mol}{L}$

The solubility product, Ksp expression for this compound is written as:

$K_s_p=[Ag^+]^2[CrO_4^2^-]$

Let's plug in the values in it:

$K_s_p=(1.5*10^-^4)^2(7.5*10^-^5)$

$K_s_p=1.69*10^-^1^2$

So, the solubility product(Ksp) of $Ag_2CrO_4$ is $1.69*10^-^1^2$ .

Answer 2

Hope this also helps