Final answer:
The concentration of K+ ions in a 0.250 M K2SO4 solution is 0.500 M, while the concentration of SO4^2- ions is 0.250 M.
Step-by-step explanation:
The question asks to determine the concentration of individual ions in a 0.250 M K2SO4 solution. This involves understanding how salts dissociate in water. Potassium sulfate (K2SO4) dissociates into two potassium ions (K+) and one sulfate ion (SO42-) per formula unit when dissolved in water.
Since every one mole of K2SO4 produces two moles of K+, the concentration of potassium ions will be 2 x 0.250 M = 0.500 M. Similarly, since every one mole of K2SO4 also produces one mole of SO42-, the sulfate ion concentration will be equal to the initial molarity of the potassium sulfate, which is 0.250 M.
In a 0.250 M K2SO4 solution, the concentration of each individual ion can be determined. Since K2SO4 dissociates completely in water, it forms two K+ ions and one SO42- ion.
Concentration of K+ ions:
Since there are two K+ ions for every one formula unit of K2SO4, the concentration of K+ ions will be twice the concentration of K2SO4.
Concentration of K+ ions = 2 * 0.250 M = 0.500 M
Concentration of SO42- ions:
The concentration of SO42- ions will be equal to the concentration of K2SO4.
Concentration of SO42- ions = 0.250 M