Question

You made up a saturated solution of calcium sulfate (CaSO4). The temperature is 25°C. You then add 5.00 × 10−3 M sodium sulfate (Na2SO4 ). What are the concentrations of calcium and sulfate after equilibrium is reached? The pKs of CaSO4 is 4.58. Answer: Ca2+ = 0.0032 M, SO2− 4 = 0.0082 M

Answer 1

Answer:

$Ca^{+2}=5.13x10^{-3}M\\SO_4^{-2}=0.010 M$

Explanation:

Hello,

At first, the answer is wrong, consider the following procedure:

- the pKs, leads to a Ks of:

$pKs=-Log(Ks)$

$Ks=10^{-4.58}=2.63x10^{-5}$

- now, stating the equilibrium for the calcium sulfate (whereas x is the change during the equilibrium):

$Ks=[Ca^{+2}]^{eq}[SO_{4}^{-2}]^{eq}\\Ks=[x]^{eq}[x]^{eq}\\Ks=x^{2}\\x=\sqrt{2.63x10^{-5}}\\x=5.13x10^{-3} M$

- x becomes the concentration of calcium ions and one of the contributions for the total concentration of sulfate ions, thus:

$[Ca^{+2}]^{eq}=5.13x10^{-3} M$

- finally, since the sodium sulfate is totally dissolved, we just simply add its concentration (equal to the sulfate ions yielded by itself) to the outgoing sulfate concentration from the calcium sulfate:

$[SO_4^{-2}]^{total}=5x10^{-3}M+5.13x10^{-3}M\\[SO_4^{-2}]^{total}=0.010 M$

Bibliography:https://books.google.com.co/books?id=LQwyfFc3pXsC&lpg=PA243&ots=zIJN3nSFWE&dq=pKs%20chemistry&pg=PA244#v=onepage&q=pKs%20chemistry&f=false page 243

Best regards!

Answer 2

Answer:

Ca^2+ =  0.0032 M

SO4^2- = 0.008205 M

Explanation:

Step 1: Data given

Temperature = 25.0 °C

Molarity of Na2SO4 = 5.00 * 10^-3 M

The pKs of CaSO4 is 4.58

Step 2: The balanced equations

CaSO4 ==> Ca^2+ + SO4^2-

Step 3:

pKsp = 4.58

Ksp = 10^-4.58 = 2.63 *10^-5

Ksp = (Ca^2+)(SO4^2-) = 2.63 *10^-5

Ksp = X * X = 2.63 *10^-5

X = 0.00513

Na2SO4 ==> 2Na^+ + SO4^2-

We add 5.00 * 10^−3 M sodium sulfate

For 1 mol Na2SO4 we have 2 moles Na+ and 1 mol SO4^2-

Molarity of Na+ = 2*5.00 * 10^-3 = 0.01 M

Molarity SO4^-2 = 5.00 * 10^-3 M = 0.005 M

Ksp = (Ca^2+)(SO4^2-)

Ksp = X*(X+0.005) = 2.63 * 10^-5

X = 0.003205

Ca^2+ = X = 0.0032 M

SO4^2- = X + 0.005 = 0.008205 M