At 25°C, the reaction CaCrO₄(𝑠) ⇌ Ca²⁺(𝑎𝑞) + CrO₄²⁻(𝑎𝑞) has an equilibrium constant 𝐾_𝑐 = 7.1×10⁻⁴. What are the equilibrium concentrations of Ca²⁺ and CrO₄²⁻ in a saturated solution of CaCrO₄?

Video transcript

Hello everyone today. We are being given the following problem and asks to solve for it. It says the equilibrium constant K. C. Of the following reaction at 25 degrees Celsius is 8.6 times 10 to the negative fifth in a saturated solution of this magnesium oxalate, calculate the the equilibrium concentrations of both magnesium and oxalic acid obsolete. So the first thing they don't want to do is we want to know our K. C. Expression So we have Casey and when right in this case the expression we're going to write the concentration of our products and the numerator over the concentration of our reactant in the denominator. And so we're also going to only want to pay attention to gasses. So with Casey expression you only pay attention to the gasses in writing this out. We're gonna start with our first reactant which is magnesium. So we're going to have MG two plus in brackets. And we're going to multiply the second reactant which is oxalate, C 2042 - also in Brackets. And since we said that we only pay attention to gasses and not solids, we are not going to write our magnesium oxalate in this case the expression and so continue on to this. Since we don't know the concentrations of magnesium or oxalate, we can simply substitute them for X. So we're going to have X times X. And then we plug in our K. C. Value of 8.6 times 10 to the negative fifth. That will equal X squared When we square both sides, we get that X. is equal to 9.3 times 10 to the negative third, and that is also going to be equal to our concentrations of magnesium two plus as well as our concentration of oxalate. I hope this helped, and until next time.