Suppose you wanted to produce an aqueous solution of pH = 8.60 by dissolving one of the following salts in water:
KNO2
At what molarity?
pH = 8.60
pOH = Kw/pH = about 6 but you need a better answer here and on the calculations below.
Convert pOH to (OH^-)
Let X = molarity of KNO2.
.........NO2%- + HOH ==> HNO2 + OH^-
I.........x...............0......0
C.........-y..............y......y
E.........x-y..............y.....y
Kb for NO2^- = Kw/Ka for HNO2 = (x)(x)/(x-y)
You know y from the OH^-. Substitute and solve for x. I'm guessing about 1 M.
Post your work if you get stuck.
To determine the molarity of the aqueous solution, we need to consider the dissociation of the salt in water and the relevant equilibrium equation. In this case, the dissociation of KNO2 in water can be represented as:
KNO2(aq) -> K+(aq) + NO2-(aq)
Since KNO2 is a salt, it is expected to dissociate completely in water, meaning that every KNO2 molecule will completely dissociate into one K+ ion and one NO2- ion.
To determine the molarity of the solution, we need to know the volume of water in which we dissolve the salt. Let's assume we are dissolving the salt in 1 liter of water.
Now, the pH of a solution depends on the concentration of hydrogen ions (H+) in the solution. To calculate the concentration of H+ ions, we can use the following equation:
pH = -log[H+]
Where [H+] represents the concentration of H+ ions.
Since we want the pH of the solution to be 8.60, we can rearrange the equation to find [H+]:
[H+] = 10^(-pH)
[H+] = 10^(-8.60)
Calculating this, we find:
[H+] = 2.51 x 10^(-9) M
Since KNO2 fully dissociates, the concentration of K+ ions will be the same as the concentration of H+ ions. Therefore, the molarity of K+ ions (and hence KNO2) in the aqueous solution will also be 2.51 x 10^(-9) M.
So, to produce an aqueous solution of pH=8.60 using KNO2, you would need to dissolve KNO2 in water with a concentration of 2.51 x 10^(-9) M.