Calculate the [OH–] in 0.01 M aqueous solution of NaOCN (Kb for OCN– = 10–10) :-10–6 M10–7 M10–8 M10–3 M
Question
Calculate the [OH–] in 0.01 M aqueous solution of NaOCN (Kb for OCN– = 10–10) :-10–6 M10–7 M10–8 M10–3 M
Solution
To calculate the [OH-] in a 0.01 M aqueous solution of NaOCN, we first need to understand that NaOCN is a basic salt and will dissociate in water to form Na+ and OCN- ions. The OCN- ion will react with water to form OH- and HOCN according to the following equilibrium:
OCN- + H2O ⇌ OH- + HOCN
The Kb expression for this reaction is:
Kb = [OH-][HOCN] / [OCN-]
We are given that the initial concentration of OCN- is 0.01 M and that Kb for OCN- is 10^-10.
Assuming that x M of OCN- reacts with water, at equilibrium we will have:
[OH-] = [HOCN] = x M [OCN-] = 0.01 - x M
Substituting these values into the Kb expression gives:
10^-10 = x^2 / (0.01 - x)
Assuming that x is much smaller than 0.01 (because Kb is very small), we can simplify this to:
10^-10 = x^2 / 0.01
Solving for x gives:
x = sqrt(10^-10 * 0.01) = 10^-6 M
Therefore, the concentration of OH- in the solution is 10^-6 M.
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