1. A 25.00 sample of 0.723M HClO4 is titrated with a 0.273M KOH solution. The H3O+ concentration after the addition of 10.0mL is___M.
2. Determine the pH of a 0.188M NH3 solution of at 25 degrees C. The Kb of NH3 is 1.76x10-5.
To solve these questions, we can use the concept of acid-base titration and the equilibrium constant, respectively. Let's address each question separately:
1. Acid-Base Titration:
In an acid-base titration, we can use the stoichiometry of the balanced equation to determine the concentration of one solution by reacting it with a known volume of the other solution.
In this case, we have a sample of 0.723 M HClO4 being titrated with a 0.273 M KOH solution. The reaction between HClO4 (acid) and KOH (base) can be balanced as follows:
HClO4 + KOH -> KClO4 + H2O
We need to find the H3O+ concentration after the addition of 10.0 mL of the KOH solution.
To solve this, we can use the concept of stoichiometry to find the number of moles of HClO4 present initially and the number of moles of KOH that react with it.
First, we need to calculate the number of moles of HClO4:
Moles of HClO4 = concentration of HClO4 x volume of HClO4 solution (in liters)
Moles of HClO4 = 0.723 M x 0.025 L = 0.018075 moles
Since the reaction between HClO4 and KOH occurs in a 1:1 ratio, this means that 0.018075 moles of KOH will react with 0.018075 moles of HClO4.
Now, we can calculate the concentration of H3O+ after the addition of 10.0 mL of the KOH solution. The volume of KOH solution added is 10.0 mL which is equal to 0.01 L.
Using the stoichiometry, we know that the balanced equation tells us that for every 1 mole of HClO4, 1 mole of H3O+ is formed. Therefore, the concentration of H3O+ can be calculated as follows:
Concentration of H3O+ = Moles of H3O+ / Volume of solution (in liters)
Concentration of H3O+ = 0.018075 moles / (0.025 L + 0.01 L) = 0.470 M
Therefore, the H3O+ concentration after the addition of 10.0 mL of the KOH solution is 0.470 M.
2. pH Calculation using Kb:
In this problem, we have a 0.188 M NH3 solution, and we need to find the pH. For this calculation, we will use the equilibrium constant Kb.
The equation for the reaction of NH3 (a weak base) with water is:
NH3 + H2O ⇌ NH4+ + OH-
The Kb expression for this reaction is:
Kb = [NH4+][OH-] / [NH3]
We are given that the Kb of NH3 is 1.76 x 10-5. Since we have the concentration of NH3, we can assume that the concentration of NH4+ and OH- are both x, as they come from the dissociation of NH3.
So, at equilibrium, the concentration of NH4+ and OH- will be x, and the concentration of NH3 will be (0.188 - x), where x is the concentration of NH4+ formed.
Using the equation for Kb and the given Kb value, we can set up the equation as follows:
(1.76 x 10^-5) = (x)(x) / (0.188 - x)
Simplifying and rearranging the equation, we get:
x^2 = (1.76 x 10^-5)(0.188 - x)
Solving this quadratic equation will give us the value of x, which represents the concentration of NH4+ (and also OH-) in the equilibrium solution.
Once we have the value of x, we can calculate the concentration of OH- in the solution. Since water undergoes autoionization, we know that [H+] = [OH-] = 1 x 10^-7 M.
Now, using the equation pH = -log10[H+], we can calculate the pH. Since [H+] = [OH-] = 1 x 10^-7 M, the pH is 7.
Therefore, the pH of the 0.188 M NH3 solution is 7.