HF + NaOH =NaF + H2O
What will be the pH of a solution produced when 40 mL of 0.25 M HF is titrated with 80 mL of 0.125M NaOH? Ka for HF is 3.5*10^-4
40 mL x 0.25M = 10.0 millimols HF.
80 mL x 0.125M = 10.0 mmols NaOH
So you are at the equivalence point and the pH will be determined by the hydrolysis of the NaF.
(NaF) = 10 mmols/120 mL - 0.0833M
..........F^- + HOH ==> HF + OH^-
init..0.0833............0.....0
change...-x.............x.....x
equil.0.0833-x...........x....x
Kb for F^- = (Kw/Ka for HF) = (HF)(OH^-)/(F^-)
Substitute the equil line from the ICE chart into the K expression and solve for x = (OH^-) then convert to pH.
See line 5.
(NaF) = 10 mmols/120 mL - 0.0833M
Replace the - with =
To find the pH after titration, we need to determine the moles of HF and NaOH used in the reaction. Then, we can calculate the concentration of the remaining species and use the dissociation constant (Ka) to find the pH.
Step 1: Calculate the moles of HF and NaOH:
Moles of HF = concentration x volume
= 0.25 M x 0.04 L (using 40 mL)
= 0.01 moles
Moles of NaOH = concentration x volume
= 0.125 M x 0.08 L (using 80 mL)
= 0.01 moles
Step 2: Determine the limiting reactant:
Since the moles of HF and NaOH used are equal, neither is in excess; both are the limiting reactants.
Step 3: Determine the moles of the remaining species:
As per the balanced equation, for every 1 mole of HF, we form 1 mole of NaF. Therefore, after the reaction, we will have 0 moles of HF remaining.
Step 4: Calculate the concentration of NaF:
The final volume of the solution is the sum of the volumes of both reactants:
Final volume = 40 mL + 80 mL
= 120 mL
= 0.12 L
Concentration of NaF = Moles of NaF / Final volume
= 0.01 moles / 0.12 L
= 0.0833 M
Step 5: Calculate the concentration of H3O+:
Since NaF is a salt, it dissociates in solution to form Na+ and F-. However, F- will react with water (H2O) to create some OH- ions.
In this reaction, 1 mole of NaF will produce 1 mole of F-, and 1 mole of H2O will produce 1 mole of OH-.
Since HF reacts with NaOH in a one-to-one stoichiometric ratio, we can say that every mole of NaF formed will produce one mole of OH-. Therefore, the concentration of OH- is equal to the concentration of NaF.
Concentration of OH- = Concentration of NaF = 0.0833 M
Now, we can use the equilibrium expression for the dissociation of water to equation:
Kw = [H3O+][OH-]
Since the concentration of OH- is known, we can rearrange the equation to solve for [H3O+]:
[H3O+] = Kw / [OH-]
The value of Kw at 25°C is 1.0 x 10^-14.
[H3O+] = (1.0 x 10^-14) / (0.0833 M)
= 1.2 x 10^-13 M
Step 6: Calculate the pH of the solution:
pH = -log10 [H3O+]
= -log10 (1.2 x 10^-13)
= 12.92
Therefore, the pH of the solution produced after titration will be approximately 12.92.
To determine the pH of a solution after a titration, we need to find the concentration of the resulting species in the solution and calculate the pH based on their equilibrium.
Step 1: Determine the limiting reactant
In this case, we are given the volumes and concentrations of both reactants. To determine the limiting reactant, we need to compare the moles of each reactant.
The moles of HF can be calculated as follows:
moles of HF = volume of HF (in L) x concentration of HF
= 0.040 L x 0.25 mol/L
= 0.010 mol
The moles of NaOH can be calculated as follows:
moles of NaOH = volume of NaOH (in L) x concentration of NaOH
= 0.080 L x 0.125 mol/L
= 0.010 mol
Since both reactants have an equal number of moles (0.010 mol), neither of them is in excess. Therefore, there is a 1:1 stoichiometric ratio between HF and NaOH in this reaction.
Step 2: Calculate the moles of the products
Since there is a 1:1 stoichiometric ratio between HF and NaOH, the moles of the products will also be equal to 0.010 mol.
Step 3: Calculate the concentration of the resulting species
Now, let's calculate the final concentration of HF and NaF in the solution after the reaction.
- HF: The initial volume of HF is 40 mL, which is the same as 0.040 L. Since the total volume of the resulting solution is 0.040 L + 0.080 L = 0.120 L, we can calculate the concentration of HF as follows:
concentration of HF = moles of HF / volume of resulting solution
= 0.010 mol / 0.120 L
= 0.083 mol/L
- NaF: The moles of NaF are also 0.010 mol, and the total volume of the resulting solution is 0.120 L. Thus, the concentration of NaF is:
concentration of NaF = moles of NaF / volume of resulting solution
= 0.010 mol / 0.120 L
= 0.083 mol/L
Step 4: Calculate the pH
To calculate the pH, we need to consider the dissociation of HF in water. HF is a weak acid, so we can use the Ka value to determine the concentration of H+ ions formed.
The dissociation of HF can be represented as follows:
HF + H2O ⇌ H3O+ + w = [H3O+][OH-] = 1.0 x 10^-14 (at 25°C)
Since HF and NaF are a conjugate acid-base pair, their concentrations can be used to calculate the concentrations of H3O+ and OH- ions.
[H3O+] = [F-] = sqrt(Ka x C(HF))
= sqrt(3.5 x 10^-4 x 0.083 mol/L)
= 1.39 x 10^-3 mol/L
Since [H3O+] = [OH-] (due to the 1:1 stoichiometric ratio between HF and NaOH), the pOH of the solution is calculated as follows:
pOH = -log[OH-]
= -log(1.39 x 10^-3)
≈ 2.86
Finally, we can calculate the pH using the pOH:
pH = 14 - pOH
= 14 - 2.86
≈ 11.14
Therefore, the pH of the solution produced when 40 mL of 0.25 M HF is titrated with 80 mL of 0.125 M NaOH is approximately 11.14.