Question

When 0.204 moles of (NH4)3PO4 is dissolved in enough water to make 1.500 L of solution, what is the total concentration of all ions present in the solution?

a) 0.136 M
b) 0.272 M
c) 0.408 M
d) 0.816 M

Answer 1

To find the total concentration of all ions present in the solution, we can calculate the concentration of (NH4)3PO4 and then consider the dissociation of the compound. The total concentration of all ions present is 0.272 M.

Step-by-step explanation:

To find the total concentration of all ions present in the solution, we need to consider the dissociation of (NH4)3PO4 in water. (NH4)3PO4 dissociates into 3 NH4+ ions and 1 PO4^3- ion. Therefore, the total concentration of all ions present is 3 times the concentration of NH4+ ions plus the concentration of PO4^3- ions.

Given that we have 0.204 moles of (NH4)3PO4 dissolved in 1.500 L of solution, we can calculate the concentration of (NH4)3PO4 as 0.204 moles / 1.500 L = 0.136 M.

Since (NH4)3PO4 dissociates into 3 NH4+ ions and 1 PO4^3- ion, the total concentration of all ions present is (3 * 0.136 M) + (1 * 0.136 M) = 0.272 M. Therefore, the correct answer is (b) 0.272 M.