A 100.0 mL sample of 0.20M HF is titrated with 0.10M KOH. Determine the pH of the solution after the addition of 100.0 mL of KOH. The Ka of HF is 3.5 x 10^-4
To determine the pH of the solution after the addition of 100.0 mL of 0.10M KOH, we can start by calculating the number of moles of HF present in the initial 100.0 mL of the 0.20M HF solution.
Step 1: Calculate the number of moles of HF in the initial solution.
Molarity (M) is defined as moles of solute per liter of solution. Therefore, the number of moles of HF in the 100.0 mL solution can be calculated as follows:
Moles of HF = Molarity x Volume
= 0.20 mol/L x 0.100 L
= 0.020 mol
Step 2: Calculate the number of moles of KOH added.
The number of moles of KOH can be calculated using the same formula:
Moles of KOH = Molarity x Volume
= 0.10 mol/L x 0.100 L
= 0.010 mol
Step 3: Determine the limiting reagent.
Since the stoichiometry between HF and KOH is 1:1, the limiting reagent is KOH because there are fewer moles of KOH (0.010 mol) compared to HF (0.020 mol).
Step 4: Calculate the excess moles of HF.
The excess moles of HF can be determined by subtracting the moles of KOH from the initial moles of HF:
Excess moles of HF = Initial moles of HF - Moles of KOH
= 0.020 mol - 0.010 mol
= 0.010 mol
Step 5: Convert the excess moles of HF to concentration.
To convert the moles of HF to concentration, divide the moles by the new volume of the solution. Since the volume of the solution is now 200.0 mL (100.0 mL of the initial HF solution + 100.0 mL of KOH), we can calculate the new concentration as follows:
Concentration of HF = Excess moles of HF / Volume of solution
= 0.010 mol / 0.200 L
= 0.050 M
Step 6: Calculate the pOH of the solution.
The pOH can be calculated using the formula:
pOH = -log10[OH-]
= -log10[Molarity of OH-]
= -log10[0.10 M]
= 1
Step 7: Calculate the pH of the solution.
Since pH + pOH = 14, we can calculate the pH value as:
pH = 14 - pOH
= 14 - 1
= 13
Thus, the pH of the solution after the addition of 100.0 mL of 0.10M KOH is 13.
To determine the pH of the solution after the addition of KOH, we need to determine the amount of excess KOH in the solution and calculate the resulting concentration of OH- ions. From there, we can calculate the pOH and then convert it to pH.
First, let's determine the amount of KOH that reacts with HF. Since the stoichiometric ratio between HF and KOH is 1:1, the number of moles of KOH that react is given by:
moles of KOH = volume of KOH added (in L) × concentration of KOH (in mol/L)
= 100.0 mL × (1 L / 1000 mL) × 0.10 mol/L
= 0.010 mol
Since the stoichiometric ratio is 1:1, the moles of HF that react is also 0.010 mol.
Now, let's determine the amount of HF remaining after the reaction. The initial amount of HF was:
moles of HF = volume of HF (in L) × concentration of HF (in mol/L)
= 100.0 mL × (1 L / 1000 mL) × 0.20 mol/L
= 0.020 mol
The moles of HF remaining after the reaction is given by:
moles of HF remaining = moles of HF initially - moles of HF that react
= 0.020 mol - 0.010 mol
= 0.010 mol
Now, we can calculate the resulting concentration of HF:
concentration of HF = moles of HF remaining / volume of solution (in L)
= 0.010 mol / (100 mL + 100 mL) × (1 L / 1000 mL)
= 0.010 mol / 0.200 L
= 0.050 mol/L
Since HF is a weak acid, we can assume that the dissociation of HF is negligible compared to the amount of HF that remains undissociated. Therefore, we can consider the concentration of HF as its initial concentration.
Now, let's calculate the concentration of OH- ions produced by the reaction between KOH and HF. Since the stoichiometric ratio is 1:1, the concentration of OH- ions is equal to the concentration of KOH that reacts:
concentration of OH- ions = moles of KOH that react / volume of solution
= 0.010 mol / 0.200 L
= 0.050 mol/L
Now, let's calculate the pOH:
pOH = -log10(concentration of OH- ions)
= -log10(0.050 mol/L)
= 1.30
Finally, let's convert pOH to pH using the relation:
pH + pOH = 14
pH = 14 - pOH
= 14 - 1.30
= 12.70
Therefore, the pH of the solution after the addition of 100.0 mL of KOH is approximately 12.70.
millimoles HF = mL x M = 100 mL x 0.2M = ??
mmoles KOH = 100 mL x 0.1M = ??
................HF + KOH ==> KF + H2O
initial........20....0.......0......0
add..................10..............
change........-10.....-10.....+10...-10
equil.........10......0........10....10
Equilibrium concns are as follows:
M = mmoles/mL
(HF) = 10/200 = ??
(KF) = 10/200
Plug into the Henderson-Hasselbalch equation and solve for pH.
OR, you can do it another way after working a number of these problems by recognizing that we are exactly half-way to the equivalence point. At that point, pH = pKa.