Solid AgCl and solid Ag2CrO4 are both in equilibrium with pure water. K_sp(AgCl) = 1.8 x 10^-10, K_sp(Ag2CrO4) = 1.4 x 10^-12. Find [Ag+] in units of 10^-4 M (two decimals; ignore hydrolysis of CrO4^2-).

JEE Advanced levelNumerical answerDifficulty 3 of 3ia070

Solid $AgCl$ and solid $Ag X_{2} CrO X_{4}$ are both in equilibrium with pure water. $K_{s p} (AgCl) = 1.8 \times 1 0^{- 10}$ , $K_{s p} (Ag X_{2} CrO X_{4}) = 1.4 \times 1 0^{- 12}$ . Find $[Ag X^{+}]$ in units of $1 0^{- 4} M$ (two decimals; ignore hydrolysis of $CrO X_{4} X^{2 -}$ ).

Numerical answer type, answer in x 10^-4 M, tolerance 0.01.

Charge balance:

[Ag X^{+}] = [Cl X^{-}] + 2 [CrO X_{4} X^{2 -}] = K_{1} / [Ag X^{+}] + 2 K_{2} / [Ag X^{+}]^{2}

. Hence

x^{3} - K_{1} x - 2 K_{2} = 0

with

K_{1} = 1.8 \times 1 0^{- 10}

2 K_{2} = 2.8 \times 1 0^{- 12}

. Solving numerically gives

x = 1.41 \times 1 0^{- 4} M

, i.e. 1.41 in units of

1 0^{- 4} M