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

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.

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IlaMends IlaMends · Answer 2 · 2019-04-04T10:06:57+02:00

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

Explanation:

[tex]SrCrO_4\rightleftharpoons Sr^{2+}+CrO_4^{2-}[/tex]

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

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

The [tex]Q_{sp}[/tex] 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 [tex]Q_{sp}[/tex] is equal to the value of [tex]K_{sp][/tex] (solubility product).

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

[tex]Q_{sp}=2.00\times 10^{-2} M\times 1.50\times 10^{-3}M=3.00\times 10^{-5}M^2[/tex]

The value of the [tex]Q_{sp}[/tex] is [tex]3.00\times 10^{-5} M^2[/tex].