Question

What is the value of q when the solution contains 2.00×10−2 m ca2+ and 3.00×10−2m cro42−? express your answer numerically?

Answer 1

6.00×10⁻²M

Step-by-step explanation:

CaCr₂O₄ is a sparingly soluble salt. The solubility product of CaCr₂O₄ is the molar concentration of Ca²⁺ times the molar concentration of Cr₂O₄²⁻

CaCr₂O₄ -> Ca²⁺ + Cr₂O₄²⁻

Q = [Ca²⁺] [Cr₂O₄²⁻]

[Ca²⁺] = 2.00×10⁻² M.

[CrO₄²⁻] = 3.00×10⁻² M.

Q = [Ca²⁺] • [CrO₄²⁻].

Q = (2.00×10⁻²)• (3.00×10⁻²) M.

Q = 6.00×10⁻²M

Answer 2

Answer is: value of Q is 6·10⁻⁴.

Chemical reaction: CaCrO₄(s) → Ca²⁺(aq) + CrO₄²⁻(aq).

[Ca²⁺] = 0,02 M.
[CrO₄²⁻] = 0,03 M.

Q = [Ca²⁺] · [CrO₄²⁻].

Q = 0,02 M · 0,03 M.

Q = 0,0006 M².

Q is the solubility product when the system is not at equilibrium.