What happens when equal volumes of 10^-5M of BaCl2 and 10^-5M of Na2SO4 are mixed at 25degree Celsius. Ksp of BaSO4=0.3×10^-10.
A ppt of BaSO4 forms.
BaCl2 + Na2SO4 ==> BaSO4(s) + 2NaCl
To determine what happens when equal volumes of 10^-5M BaCl2 and 10^-5M Na2SO4 are mixed, we need to consider whether a precipitate will form. This can be determined by comparing the ion concentrations of the ions that combine to form the potential precipitate, in this case, BaSO4.
First, let's start by writing the balanced equation for the reaction between BaCl2 and Na2SO4:
BaCl2(aq) + Na2SO4(aq) → BaSO4(s) + 2NaCl(aq)
From this equation, we can see that 1 mole of BaCl2 reacts with 1 mole of Na2SO4 to produce 1 mole of BaSO4 and 2 moles of NaCl.
Since we are given that the initial concentrations of both BaCl2 and Na2SO4 are 10^-5M, this means that each solution contains 10^-5 moles of the respective compound per liter of solution.
When the solutions are mixed, they will mix uniformly, and the volumes will double. Therefore, we can assume that the final volume is 2 liters. Consequently, the concentration of each compound after mixing is halved.
Now, let's calculate the concentrations of Ba2+ and SO4^2- ions after mixing:
[Ba^2+] = [BaCl2] / 2
= 10^-5M / 2
= 5 × 10^-6M
[SO4^2-] = [Na2SO4] / 2
= 10^-5M / 2
= 5 × 10^-6M
Now, let's compare the ion product (Qsp) with the solubility product constant (Ksp) value for BaSO4.
Qsp = [Ba^2+][SO4^2-]
= (5 × 10^-6M) × (5 × 10^-6M)
= 25 × 10^-12
Comparing the value of Qsp = 25 × 10^-12 with the given Ksp = 0.3 × 10^-10, we find that Qsp is smaller than Ksp. This means that no precipitation will occur because the ion product is smaller than the solubility product.
Therefore, when equal volumes of 10^-5M BaCl2 and 10^-5M Na2SO4 are mixed at 25 degrees Celsius, no precipitate of BaSO4 will form.