Question

Consider these compounds:

Answer 1

Answer:

Without doing any calculations it is possible to determine that silver chromate is more soluble than C,D , and silver chromate is less soluble than none .

It is not possible to determine whether silver chromate is more or less soluble than A,B by simply comparing Kspvalues

Explanation:

For this kind of question you need to obtain the balanced equation for the solubility, first, you will write the dissociation equation for every molecule, obtaining the ions in which it dissociates. Then you substitute in the Kps equation

$A_{n}B_{m} -> nA^{m+}+mB^{n-} \\K_{s}=[A^{m}]^{n}[B^{n}]^{m}$

Considering the corresponding stoichiometry you'll substitute in your dissociation equations as follow.

$Ag_{2}CrO_{4}(s)->2Ag^{+}(aq)+CrO_{4} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\A) PbF_{2}(s)->Pb^{2+} (aq)+2F^{-}\\s->s+2s\\s->[s][2s]^{2}=4s^{3}\\\\\\B)Ag_{2}SO_{3}(s)->2Ag^{+}(aq)+SO_{3} ^{2-}\\ s ->2s+s\\s->[2s]^{2}[s]=4s^{3} \\\\\\C)NiCO_{3}(s)->Ni^{2+}(aq)+CO_{3} ^{2-}\\ s ->s+s\\s->[s][s]=s^{2} \\\\\\D)AgCl(s)->Ag^{+}(aq)+Cl^{-}(aq)\\ s ->s+s\\s->[s][s]=s^{2}$

IIn this case, the silver chromate has a Kps of $4sx^{3}$, same as the compound in options A and B, comparing these numbers you can't determine which one is bigger. Finally, options C and D have a kps of $s^{2}$, this value is smaller than silver chromate's kps.

I hope you find this information useful! good luck!