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:The value of the
`Q_(sp)` is
`3.00* 10^(-5) M^2`.

Step-by-step explanation:

`SrCrO_4\rightleftharpoons Sr^(2+)+CrO_4^(2-)`

`[Sr^(2+)]=2.00* 10^(-2) M`

`[CrO_4^(2-)]=1.50* 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* [CrO_4^(2-)]^1`

`Q_(sp)=2.00* 10^(-2) M* 1.50* 10^(-3)M=3.00* 10^(-5)M^2`

The value of the
`Q_(sp)` is
`3.00* 10^(-5) M^2`.

Answer 2

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.