Consider the following reaction.

CaSO4(s) reverse reaction arrow Ca2+(aq) + SO42-(aq)
At 25°C the equilibrium constant is Kc = 2.4x10^-5 for this reaction.

(a) If excess CaSO4(s) is mixed with water at 25°C to produce a saturated solution of CaSO4, what are the equilibrium concentrations of Ca2+ and SO42-?

(b) If the resulting solution has a volume of 1.4 L, what is the minimum mass of CaSO4(s) needed to achieve equilibrium?

For a I got 4.9*10^5 M for both concentrations. I'm not sure how to do b. Can someone show me what steps to do?

To answer part (b) of the question, we need to determine the minimum mass of CaSO4(s) needed to achieve equilibrium in a solution with a volume of 1.4 L.

To start, let's define the given equilibrium constant expression:

Kc = [Ca2+][SO42-] / [CaSO4]

Since the concentration of a solid (CaSO4) does not change, its concentration remains constant and does not affect the equilibrium constant. Therefore, we can ignore [CaSO4] in the equilibrium constant expression.

Now, we can rearrange the equation to solve for [CaSO4]:

[CaSO4] = [Ca2+][SO42-] / Kc

Given that the equilibrium concentrations of Ca2+ and SO42- are both 4.9x10^5 M (as determined in part (a)), and the equilibrium constant (Kc) is 2.4x10^-5, we can substitute these values into the equation:

[CaSO4] = (4.9x10^5)(4.9x10^5) / 2.4x10^-5

Simplifying this expression, we get:

[CaSO4] = (4.9x10^5)^2 / 2.4x10^-5

To evaluate this expression, we can square 4.9x10^5 and divide it by 2.4x10^-5, which gives:

[CaSO4] = 9.601x10^9 M

However, the concentration of a solid is typically expressed in terms of mass per unit volume. Therefore, we need to convert the concentration to mass.

Recall that the molar mass of CaSO4 is 136.14 g/mol. To convert the concentration into mass, we need to multiply the molar mass by the number of moles in the given volume.

First, we need to convert the volume from liters to milliliters:

1.4 L = 1.4 x 1000 mL = 1400 mL

Next, we can use the equation:

mass = concentration x volume

mass = 9.601x10^9 M x 1400 mL

The unit for mass will be grams, which is equal to 1 mL of a pure substance.

Thus,

mass = 9.601x10^9 x 1400 grams

Evaluating this expression, we get:

mass = 1.344x10^13 grams

Therefore, the minimum mass of CaSO4(s) needed to achieve equilibrium in a 1.4 L solution is approximately 1.344x10^13 grams.

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