Fundamental G21

[Ahjin Kim 2B]

Can someone explain how to get the answer for this question?

A solution is prepared by dissolving 0.500 g of KCl, 0.500 g of K2S, and 0.500 g of K3PO4 in 500. mL of water. What is the concentration in the final solution of (a) potassium ions; (b) sulfide ions?

[Seung Won Annie Lee]

The final answer asks us to find the concentration of K (Molarity) in 500ml of water.

Molarity of K= mol k/ volume

First we need to find the total mol of K. Since given the mass of K in different solutes, we can find mol of K by (given mass/ molar mass of solute)x ( n mol K/ 1 mol of solute)

for example, for K2S. (0.500gK2S/110.26 g/mol) x (2 mol K/ 1 mol K2S)

we find the mol of K for each solutes (KCI, K2S, K3PO4). add each mol of K to find total mol of K

total mol of K= 2.29 x 10^-2 mol K


to find the concentration of K, use the equation above (Molarity of K= mol k/ volume); convert 500ml to l
Molarity of k= (2.29 x 10^-2 mol K/ 0.5L)

Molarity of K or concentration = 4.58x10^-2 M