Answer:
[tex]Ca^{+2}=5.13x10^{-3}M\\SO_4^{-2}=0.010 M[/tex]
Explanation:
Hello,
At first, the answer is wrong, consider the following procedure:
- the pKs, leads to a Ks of:
[tex]pKs=-Log(Ks)[/tex]
[tex]Ks=10^{-4.58}=2.63x10^{-5}[/tex]
- now, stating the equilibrium for the calcium sulfate (whereas x is the change during the equilibrium):
[tex]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[/tex]
- x becomes the concentration of calcium ions and one of the contributions for the total concentration of sulfate ions, thus:
[tex][Ca^{+2}]^{eq}=5.13x10^{-3} M[/tex]
- 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:
[tex][SO_4^{-2}]^{total}=5x10^{-3}M+5.13x10^{-3}M\\[SO_4^{-2}]^{total}=0.010 M[/tex]
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:
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
