PROBLEM 109P

A solution contains an unknown amount of dissolved calcium. Addition of 0.112 mol of K3PO4 causes complete precipitation of all of the calcium. How many moles of calcium were dissolved in the solution? What mass of calcium was dissolved in the solution?

Solution 109P

Here, we have to calculate the dissolved mass and mole of calcium in the solution.

The balanced chemical equation for the reaction is

3Ca2+(aq) + 2K3PO4(s)--------> Ca3(PO4)2 + 6K+

Thus, 0.112 mol of K3PO4 contains 0.112 mole of (PO42-

Since, from the above equation, it is found that

3 mol of Ca2+ is consumed by 2 mol of K3PO4.

Therefore the required Ca is = = 0.168 mol

Mass of Ca required = 40.078 g/mol x 0.168 mol =6.73 g

Therefore, 0.168 moles of calcium were dissolved in the solution.

6.73 g of calcium were dissolved in the solution.