Question

What is the value of q when the solution contains 2.00×10−2 m sr2+ and 1.50×10−3m cro42−?

Answer 1

Answer is: ion product for strontium chromate is 3·10⁻⁵.
Balanced chemical reaction (dissociation) of strontium chromate:
Sr²⁺(aq) + CrO₄²⁻(aq) → SrCrO₄.
Qsp(SrCrO₄) = c(Sr²⁺)·(CrO₄²⁻).
c(Sr²⁺) = 2.00·10⁻² M.
c(CrO₄²⁻) = 1.50·10⁻³ M.
Q = 2.00·10⁻² M · 1.50·10⁻³ M.
Q = 0.00003 = 3·10⁻⁵ M².

Answer 2

Answer:The value of the $Q_{sp}$ is $3.00\times 10^{-5} M^2$.

Explanation:

$SrCrO_4\rightleftharpoons Sr^{2+}+CrO_4^{2-}$

$[Sr^{2+}]=2.00\times 10^{-2} M$

$[CrO_4^{2-}]=1.50\times 10^{-3}M$

The $Q_{sp}$ of the salt solution is defined as the product of of the concentration of the ions raised to power equal to their stoichiometric coefficient.

At equilibrium the value of $Q_{sp}$ is equal to the value of $K_{sp]$ (solubility product).

$Q_{sp}=[Sr^{2+}]^1\times [CrO_4^{2-}]^1$

$Q_{sp}=2.00\times 10^{-2} M\times 1.50\times 10^{-3}M=3.00\times 10^{-5}M^2$

The value of the $Q_{sp}$ is $3.00\times 10^{-5} M^2$.