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?
Use the HH equation.
pH = pKa + log(b/a)
1.00 = pKa + log(b/0.025)
Calculate pKa and solve for (base).
Well, I must say, this is a rather acidic situation. A pH of 1.00 is definitely in the "ouch" range. But let's not dwell on the acidity, let's crunch some numbers instead!
First, we need to calculate the concentration of H+ ions in the solution. Since the pH is 1.00, we know that [H+] = 10^(-pH). So, [H+] = 10^(-1.00) = 0.1 M.
Now, let's use the equation for the ionization of HIO3: HIO3 -> H+ + IO3-. The Ka value given tells us that Ka = [H+][IO3-]/[HIO3]. Since we know the concentration of H+ ions (0.1 M) and the concentration of HIO3 (0.025 M), we can rearrange the equation to find [IO3-].
Ka = [H+][IO3-]/[HIO3]
1.7x10^(-1) = (0.1 M)[IO3-]/(0.025 M)
Simplifying the equation, we get:
1.7x10^(-1) = 4[IO3-]
[IO3-] = (1.7x10^(-1))/4 = 4.25x10^(-2) M
So, the molarity of KIO3 is 4.25x10^(-2) M. Remember, always be careful with those acids, folks!
To find the molarity of KIO3, we need to first write the balanced chemical equation for the reaction between HIO3 and KIO3 when dissolved in water.
HIO3 + KIO3 → H+ + IO3- + K+
Since the concentration of H+ ions in the solution will determine the pH, we can use the dissociation constant (Ka) of HIO3 to find the concentration of H+ ions.
Ka = [H+][IO3-] / [HIO3]
Given that Ka = 1.7x10^-1, and the molarity of HIO3 is 0.025 M, we can substitute these values into the equation and solve for the concentration of H+ ions ([H+]).
1.7x10^-1 = [H+][IO3-] / 0.025
Now, we need to determine the concentration of IO3- ions. Since HIO3 completely dissociates in water, the concentration of IO3- ions will be the same as the concentration of HIO3, which is 0.025 M.
1.7x10^-1 = [H+][0.025] / 0.025
Simplifying the equation, we get:
1.7x10^-1 = [H+]
Since the pH of the solution is given as 1.00, we can use the equation pH = -log[H+] to find the value of [H+].
1.00 = -log[H+]
Taking the antilog of both sides, we get:
[H+] = 10^-1.00
[H+] = 0.1 M
Now, we know that [H+] = [KIO3] (since HIO3 and KIO3 are in a 1:1 molar ratio), so the molarity of KIO3 is 0.1 M.
To solve this problem, we need to use the dissociation of HIO3 and the concept of the equilibrium constant to find the molarity of KIO3. Let's go step by step:
Step 1: Write the balanced chemical equation for the dissociation of HIO3 in water:
HIO3 ⇌ H+ + IO3-
Step 2: Write the expression for the equilibrium constant (Ka) for the dissociation reaction:
Ka = [H+][IO3-] / [HIO3]
Step 3: Given that the value of Ka is 1.7x10^-1 and the molarity of HIO3 is 0.025 M, we can set up the equation:
1.7x10^-1 = [H+][IO3-] / 0.025
Step 4: Since the pH of the solution is 1.00, we know that the concentration of H+ is 10^-pH:
[H+] = 10^-1 = 0.1 M
Step 5: Substitute the concentration of H+ into the equation:
1.7x10^-1 = 0.1[IO3-] / 0.025
Step 6: Rearrange the equation to solve for [IO3-]:
[IO3-] = (1.7x10^-1)(0.025) / 0.1
Step 7: Calculate the value of [IO3-]:
[IO3-] = 4.25x10^-3 M
Therefore, the molarity of KIO3 is 4.25x10^-3 M.