the pH of a solution of HIO3 (Ka=1.7x10^-1) and KIO3 is 1.00. what is the molarity of KIO3 if the molarity of HIO3 is 0.025 M?
See your post above.
To find the molarity of KIO3 in the given solution, we need to set up an equation using the equilibrium constant expression for the dissociation of HIO3. The given Ka value for HIO3 is 1.7x10^-1.
First, let's write the dissociation equation for HIO3:
HIO3 ⇌ H+ + IO3-
The initial concentration of HIO3 is given as 0.025 M. Let's assume the change in the concentration of H+ is x M. Since HIO3 dissociates in a 1:1 ratio with H+, the concentration of IO3- will also be x M.
Using the equation, we can set up the equilibrium constant expression:
Ka = [H+][IO3-] / [HIO3]
Substituting the given values, we have:
1.7x10^-1 = x * x / 0.025
Simplifying the equation further:
1.7x10^-1 = x^2 / 0.025
Cross-multiplying and rearranging the equation:
x^2 = 1.7x10^-1 * 0.025
x^2 = 4.25x10^-3
Taking the square root of both sides:
x ≈ 6.52x10^-2
Since x represents the concentration of H+ ions, it also represents the concentration of IO3- ions and KIO3 since they all have the same molar ratio.
Therefore, the molarity of KIO3 in the solution is approximately 6.52x10^-2 M.