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GaryK [48]
4 years ago
5

At 25 ∘C the reaction CaCrO4(s)←→Ca2+(aq)+CrO2−4(aq) has an equilibrium constant Kc=7.1×10−4. What is the equilibrium concentrat

ion of Ca2+ in a saturated solution of CaCrO4?
Chemistry
1 answer:
Nitella [24]4 years ago
6 0

Answer:

2.67 × 10⁻²

Explanation:

Equation for the reaction is expressed as:

CaCrO₄(s)    ⇄      Ca₂⁺(aq)         +        CrO₂⁻⁴(aq)

Given that:

Kc=7.1×10⁻⁴

Kc= [Ca^{2+}][CrO^{2-}_4]

Kc= [x][x]

Kc= [x²]

7.1×10⁻⁴ =  [x²]

x = \sqrt{7.1*10^{-4}}

x = 0.0267

x = 2.67*10^{-2}

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