Calculate the molar concentration of OH− ions in an 1.79 M solution of hypobromite ion (BrO−; Kb = 4.0 10-6).
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To calculate the molar concentration of OH- ions in a solution of hypobromite ion (BrO-), we need to use the Kb value and the given concentration of hypobromite ion.
First, let's write the chemical equation for the reaction of hypobromite ion with water:
BrO- + H2O ⇌ OH- + HOBr
We can see that in this reaction, BrO- functions as a base and accepts a proton (H+) to form OH-. The Kb value (the base dissociation constant) represents the extent to which the base will dissociate and produce OH- ions.
The expression for Kb is:
Kb = [OH-] * [HOBr] / [BrO-]
We are given the Kb value as 4.0 x 10^-6. Now, let's assume the concentration of OH- ions at equilibrium is x M. Since the concentration of OH- ions is the same as the concentration of the products in the equilibrium expression, we can substitute [OH-] with x in the expression and rewrite it as:
Kb = x * [HOBr] / [BrO-]
Next, we need to determine the concentrations of HOBr and BrO-. The only concentration given in the problem is that of BrO-, which is 1.79 M. We'll use this concentration to calculate the concentration of HOBr.
Since HOBr and BrO- are formed by the reaction of hypobromite ion with water, we can assume that the initial concentration of HOBr is also x M, as OH- and HOBr are both products of this reaction. Therefore, the concentration of HOBr is also x M.
Finally, we can plug in the values into the expression for Kb:
4.0 x 10^-6 = x * x / 1.79
Now we can solve for x, which is the concentration of OH- ions:
x^2 = 4.0 x 10^-6 * 1.79
x^2 = 7.16 x 10^-6
Taking the square root of both sides:
x ≈ 8.46 x 10^-4 M
Therefore, the molar concentration of OH- ions in the 1.79 M solution of hypobromite ion is approximately 8.46 x 10^-4 M.