you made up a standard solution of calcium sulfate (CaSO4). the temperature is 25C. you then add 5.00*10^-3 odium sulfate (Na2SO4). Calculate the concentrations of calcium and sulfate after equilibrium is reached? The pKs of CaSO4 is 4.58.
CaSO4 ==> Ca^+2 SO4^-2
Ksp = (Ca^+2)(SO4^-2) = 4.58E?? (you had better look up the Ksp because 4.58 can't possibly be correct.)
You then add 5.00*10^-3 WHAT of Na2SO4.
Take care of the errors in this and I'll get back to it.
the pKs is correct. If you 10 to the negative power of pKs you get the Ks. I'm having a problem witht this problem as well. Any help? And yes, you add 5*10^-3 of Na2SO4.
To calculate the concentrations of calcium and sulfate after equilibrium is reached, we need to consider the solubility product constant (Ksp) for calcium sulfate (CaSO4). The Ksp expression for CaSO4 is given by:
Ksp = [Ca2+][SO42-]
Since the pKs of CaSO4 is given as 4.58, we can convert it to Ksp by using the equation:
Ksp = 10^(-pKs)
Ksp = 10^(-4.58)
Now, let's denote the initial concentration of calcium sulfate as [Ca2+]0 and [SO42-]0. Since we made up a standard solution, the initial concentrations of both Ca2+ and SO42- are equal. Therefore, [Ca2+]0 = [SO42-]0 = x (assuming the concentration in moles per liter, M).
After equilibrium is reached, some of the calcium sulfate will dissolve and dissociate into Ca2+ and SO42-. Let's say the amount of calcium sulfate that dissolved is 'y' (in moles per liter, M). As a result, the concentrations of Ca2+ and SO42- will increase by 'y'. Therefore, the final concentrations of Ca2+ and SO42- are:
[Ca2+] = [Ca2+]0 + y
[SO42-] = [SO42-]0 + y
Using the solubility product constant equation, we know that:
Ksp = [Ca2+][SO42-]
Plugging in the concentrations, we get:
Ksp = ([Ca2+]0 + y)([SO42-]0 + y)
Now we can substitute [Ca2+]0 = [SO42-]0 = x and solve for y:
Ksp = (x + y)(x + y)
10^(-4.58) = (x + y)^2
Taking the square root of both sides:
(x + y) = √(10^(-4.58))
Next, we know that we added 5.00*10^-3 moles of sodium sulfate (Na2SO4) to the solution. Since each mole of Na2SO4 dissociates into 1 mole of SO42-, the amount of SO42- that formed in moles is equal to 5.00*10^-3. Therefore, 'y' is equal to 5.00*10^-3.
Substituting this value back into the equation (x + y) = √(10^(-4.58)), we can solve for 'x':
(x + 5.00*10^-3) = √(10^(-4.58))
Solving for 'x', we subtract 5.00*10^-3 from both sides:
x = √(10^(-4.58)) - 5.00*10^-3
Finally, we can calculate the concentrations of calcium and sulfate after equilibrium is reached as [Ca2+] = x and [SO42-] = x.