What will the molar concentration of K⁺ aqueous be if 29 grams of KF solid with a molar mass of 58 grams per mole is dissolved in enough water to make 2.00 liters of solution? A) 0.010 molar, B) 0.050 molar, C) 0.25 molar, D) 0.50 molar, or E) 1.0 molar.
In this question, we’re adding solid KF, or potassium fluoride, to water so that it dissolves to form K⁺ and F⁻ ions. We’re tasked with finding the molar concentration of the K⁺ ions, which is the moles of the K⁺ ions divided by the liters of solution. We’re given the mass of potassium fluoride that we started with and the volume of the solution. So we need to find the moles of potassium ions in order to find the concentration.
We start with 29 grams of potassium fluoride. And we need to convert this into moles of potassium fluoride. We can do this by dividing by the molar mass of potassium fluoride. Now we’re in moles of potassium fluoride. But we want moles of K⁺. According to our balanced chemical equation, there’s one mole of potassium fluoride for every one mole of K⁺. This gives us 0.5 moles of potassium ions.
If we wanted to, we could do this calculation in two steps. First finding the amount of potassium fluoride in moles by dividing the mass by the molar mass and then using the mole ratio to convert from moles of potassium fluoride to moles of potassium ions. Other way, we can now find the concentration of the potassium ions by dividing the moles of the potassium ions by the volume of the solution, which is two liters. This gives us 0.25 molar, which matches answer choice C. So the molar concentration of the potassium ions from dissolving our potassium fluoride in water is 0.25 molar.