What is the concentration of K+ in 0.15 M K2S?
The concentration of K+ is 0.30 M. Potassium sulfide (K2S) dissociates completely in water into 2 K+ ions and 1 S2- ion per formula unit, so [K+] = 2 x 0.15 M = 0.30 M.
The answer
The concentration of potassium ion is 0.30 M.
Potassium sulfide is a soluble ionic compound, so in water it dissociates completely:
K2S -> 2 K+ + S2-
The subscript 2 on potassium means every formula unit of K2S releases two potassium ions and one sulfide ion. Because the salt fully dissolves, the potassium concentration is twice the salt concentration:
[K+] = 2 x 0.15 M = 0.30 M
How the stoichiometric ratio works
The key idea is the mole ratio inside the balanced dissociation equation. For any strong electrolyte, multiply the molarity of the compound by the number of that ion in one formula unit:
- K+ : K2S ratio is 2 : 1, so [K+] = 2 x 0.15 = 0.30 M
- S2- : K2S ratio is 1 : 1, so [S2-] = 1 x 0.15 = 0.15 M
This is why simply reading the answer as "0.15 M" is wrong — that ignores the subscript. A common mistake is to assume every dissolved salt gives ion concentrations equal to the salt concentration. That is only true when the ion has a subscript of 1 (like S2- here). Whenever a subscript is 2, 3, or more, the ion concentration scales up accordingly.
Ruling out the wrong values
- 0.15 M — this would be correct only if K2S produced one K+ per formula unit. It produces two, so this undercounts by half.
- 0.075 M — this comes from dividing instead of multiplying (0.15 ÷ 2). Dissociation adds ions; it never dilutes a specific ion below the parent molarity.
- 0.45 M — this would require three K+ per formula unit (as in K3PO4), but K2S has only two.
The bigger picture
This calculation is the foundation of solution stoichiometry and shows up in colligative properties, conductivity, and precipitation problems. The general rule: for a salt that dissolves as AxBy -> x A^n+ + y B^m-, the concentration of A is x times the salt molarity and the concentration of B is y times the salt molarity. Apply it to a few examples to lock it in:
- 0.10 M NaCl -> [Na+] = 0.10 M, [Cl-] = 0.10 M (1:1)
- 0.10 M CaCl2 -> [Ca2+] = 0.10 M, [Cl-] = 0.20 M (1:2)
- 0.10 M Al2(SO4)3 -> [Al3+] = 0.20 M, [SO4 2-] = 0.30 M
For K2S specifically, the two potassium ions make [K+] = 0.30 M, while [S2-] stays at 0.15 M.
Frequently asked
How do you find ion concentration from molarity?
Write the balanced dissociation equation, then multiply the salt's molarity by the number of that ion in one formula unit. For a soluble salt AxBy, [A] equals x times the molarity and [B] equals y times the molarity.
How many potassium ions are in K2S?
There are two potassium ions per formula unit of K2S. The subscript 2 on K means each formula unit releases 2 K+ ions along with 1 sulfide (S2-) ion when it dissolves.
What is the concentration of S2- in 0.15 M K2S?
The sulfide concentration is 0.15 M. There is only one S2- ion per formula unit (subscript 1), so [S2-] equals the salt concentration of 0.15 M, while [K+] is doubled to 0.30 M.
How does dissociation affect ion molarity?
Dissociation multiplies each ion's concentration by its subscript in the formula. It never divides or lowers an ion below the salt molarity, so ions with subscripts greater than one end up more concentrated than the dissolved compound.