Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.140 M HClO(aq) with 0.140 M KOH(aq).HClO is a weak acid with a Ka of 4.0× 10–8. It reacts with strong base to produce ClO–.
(e) after addition of 60.0 mL of KOH
HClO + KOH ==> KClO + H2O
mols HClO = M x L = 0.140 x 0.0500 = about 0.007
mols KOH = 0.0600 x 0.14 = 0.0084
This is past the equivalence point.
mols excess KOH = 0.0084-0.007 = 0.0014
M (OH^-) = 0.0014/0.110 L = ? and convert to pOH, then to pH.
To calculate the pH after the addition of 60.0 mL of KOH to the 50.0 mL of 0.140 M HClO solution, we need to determine the number of moles of HClO and KOH reacted and the resulting concentration of the remaining HClO and ClO- species.
Here's the step-by-step process to calculate the pH:
Step 1: Determine the initial moles of HClO in the 50.0 mL solution.
- Moles of HClO = (Volume of HClO solution in L) x (Molarity of HClO)
- Volume of HClO solution = 50.0 mL = 0.050 L
- Molarity of HClO = 0.140 M
- Moles of HClO = 0.050 L x 0.140 M = 0.007 moles
Step 2: Determine the initial moles of KOH in the 60.0 mL solution.
- Moles of KOH = (Volume of KOH solution in L) x (Molarity of KOH)
- Volume of KOH solution = 60.0 mL = 0.060 L
- Molarity of KOH = 0.140 M
- Moles of KOH = 0.060 L x 0.140 M = 0.0084 moles
Step 3: Determine the limiting reagent and the moles of excess reagent.
- The balanced equation for the reaction is:
HClO + KOH -> KClO + H2O
- According to the stoichiometry of the reaction, the mole ratio of HClO to KOH is 1:1.
- Since the moles of KOH (0.0084 moles) are greater than the moles of HClO (0.007 moles), KOH is the limiting reagent.
- The excess moles of KOH are calculated as:
Excess moles of KOH = Moles of KOH - Moles of HClO
= 0.0084 moles - 0.007 moles
= 0.0014 moles
Step 4: Determine the moles of HClO and ClO- remaining after the reaction.
- Moles of HClO remaining = 0 moles (since it is completely reacted)
- Moles of ClO- produced = Moles of KOH reacted = Moles of KOH - Excess moles of KOH
= 0.0084 moles - 0.0014 moles
= 0.007 moles
Step 5: Calculate the concentration of HClO and ClO- remaining.
- Concentration of HClO remaining = (Moles of HClO remaining) / (Total volume of solution)
= 0 moles / (0.050 L + 0.060 L)
= 0 M (since all HClO reacted)
- Concentration of ClO- remaining = (Moles of ClO- remaining) / (Total volume of solution)
= 0.007 moles / (0.050 L + 0.060 L)
= 0.0586 M
Step 6: Calculate the pOH of the remaining ClO- solution.
- pOH = -log10(concentration of OH-)
- Since ClO- reacts with the strong base KOH, it forms OH- ions.
- Concentration of OH- = Concentration of ClO-
= 0.0586 M
- pOH = -log10(0.0586) = 1.23
Step 7: Calculate the pH from pOH.
- The pH is calculated using the expression pH = 14 - pOH.
- pH = 14 - 1.23 = 12.77
Therefore, the pH after the addition of 60.0 mL of KOH is approximately 12.77.
To calculate the pH after the addition of 60.0 mL of KOH, we need to determine the amount of HClO and KOH that reacted and the resulting concentration of OH- ions.
Step 1: Calculate the number of moles of HClO initially present:
n(HClO) = (volume)(concentration)
n(HClO) = (0.050 L)(0.140 M) = 0.007 moles
Step 2: Calculate the number of moles of KOH added:
n(KOH) = (volume)(concentration)
n(KOH) = (0.060 L)(0.140 M) = 0.0084 moles
Step 3: Determine the limiting reactant:
The stoichiometry of the reaction between HClO and KOH is 1:1, so the limiting reactant is the one with fewer moles. In this case, it is the HClO with 0.007 moles.
Step 4: Determine the moles of HClO left after the reaction:
The reaction consumes all the moles of the limiting reactant, so there are zero moles of HClO left.
Step 5: Determine the moles of ClO- formed:
The stoichiometry of the reaction tells us that for every mole of HClO consumed, one mole of ClO- is formed. Therefore, there are 0.007 moles of ClO- formed.
Step 6: Determine the concentration of ClO-:
The total volume of the solution after adding 60.0 mL of KOH is (50.0 mL + 60.0 mL) = 0.110 L. Therefore, the concentration of ClO- is:
[ClO-] = (moles)/(volume) = (0.007 moles)/(0.110 L) = 0.0636 M
Step 7: Calculate the concentration of OH- ions:
Since the KOH is a strong base, it completely dissociates in water to produce OH- ions. Therefore, the concentration of OH- ions is also 0.0636 M.
Step 8: Calculate the pOH:
pOH = -log10[OH-]
pOH = -log10(0.0636) = 1.196
Step 9: Calculate the pH:
Since pH + pOH = 14, we can use this relationship to find the pH:
pH = 14 - pOH
pH = 14 - 1.196 = 12.804
Therefore, the pH after the addition of 60.0 mL of KOH is approximately 12.804.