Find the pH at each of the following points in the titration of 25 ml of 3.0M HF with 0.3 M NaOH. The Ka value is 6.6x10-4. After adding 25 ml of 0.3M NaOH

HF + NaOH ==> NaF + H2O

millimols HF initially = mL x M = 25 x 3 = 75
mmols NaOH added = 25 x 0.3 = 7.5
mmols HF left un-neutralized = 75-7.5 = 67.5
volume of new solution = 26 mL + 25 mL = 50 mL.
M HF left = mmol/mL = 67.5/50 = 1.35 M.

......HF ==> H^+ + F^-
I....1.35.....0.....0
C.....-x......x.....x
E....1.5-x....x.....x

Substitute the E line into Ka expression for HF and solve for x = (H^+), then convert to pH.

To find the pH at each point in the titration, we need to determine the moles of acid and moles of base at each stage and then use the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

First, let's calculate the initial moles of acid and base before any reaction occurs:

Moles of HF = (Concentration of HF) x (Volume of HF)
= (3.0 mol/L) x (0.025 L)
= 0.075 mol

Moles of NaOH = (Concentration of NaOH) x (Volume of NaOH)
= (0.3 mol/L) x (0.025 L)
= 0.0075 mol

At this stage, all the HF present will react with the NaOH, following the balanced equation:

HF + NaOH -> NaF + H2O

Since the stoichiometry is 1:1, the moles of the excess reactant will be:

Moles of excess NaOH = Moles of NaOH - Moles of HF
= 0.0075 mol - 0.075 mol
= -0.0675 mol (excess NaOH)

The negative value indicates a deficit of HF, meaning that all the HF has reacted.

For this point in the titration, the moles of the conjugate base (F-) will be equal to the moles of NaOH used, and the moles of the acid (HF) will be zero.

Now, let's use the Henderson-Hasselbalch equation with the given Ka value (6.6x10^-4) to calculate the pH:

pH = pKa + log([A-]/[HA])

At the equivalence point, [A-]/[HA] = 1, as we have equal moles of base and acid.

pH = pKa + log(1)
pH = pKa

pH = -log(Ka)
pH = -log(6.6x10^-4)
pH ≈ 3.18

Therefore, after adding 25 ml of 0.3 M NaOH to 25 ml of 3.0 M HF, the pH of the solution would be approximately 3.18.

To find the pH at each point in the titration, we need to consider the reaction between the strong base NaOH and the weak acid HF. This is an example of an acid-base titration.

First, let's calculate the moles of HF initially present in 25 mL of 3.0 M HF solution:

Moles of HF = Volume (L) x Concentration (M)
= 25 mL x (1 L / 1000 mL) x 3.0 M
= 0.075 moles of HF

Since the stoichiometric ratio between HF and NaOH is 1:1, it means that it takes 0.075 moles of NaOH to completely neutralize 0.075 moles of HF.

At the beginning of the titration, before any NaOH is added, the moles of HF remaining will be equal to the moles initially present:

Moles of HF remaining = 0.075 moles

Next, let's calculate the moles of NaOH added:

Moles of NaOH = Volume (L) x Concentration (M)
= 25 mL x (1 L / 1000 mL) x 0.3 M
= 0.0075 moles of NaOH

Now, let's consider the reaction between HF and NaOH:

HF + NaOH → NaF + H2O

Since NaF is a salt and completely dissociates in water, we will have a solution containing NaF and water after the reaction has taken place.

To determine the pH of the resulting solution, we need to calculate the moles of excess NaOH or moles of HF remaining after reaction:

Moles of HF remaining = Moles of HF initially present - Moles of NaOH added
= 0.075 moles - 0.0075 moles
= 0.0675 moles of HF

Now, let's calculate the concentration of HF after the reaction:

Volume of HF after reaction = 25 mL + 25 mL = 50 mL

Concentration of HF = (Moles of HF remaining) / (Volume of HF after reaction in L)
= 0.0675 moles / (50 mL x 1 L / 1000 mL)
= 1.35 M

Since NaF is a salt that dissolves completely in water, we can assume that the resulting solution is a neutral solution. Therefore, to calculate the pH, we can calculate the pOH and then convert it to pH using the equation:

pOH = -log10 (OH- concentration)

The concentration of OH- ions can be calculated using the equation:

OH- concentration = (Moles of excess NaOH) / (Volume of solution in L)

Moles of excess NaOH = Moles of NaOH added - Moles of HF remaining
= 0.0075 moles - 0.0675 moles
= -0.06 moles (negative because it's in excess)

Since we have negative moles of excess NaOH, it means that all the excess NaOH has been neutralized by HF, resulting in a slightly acidic solution.

Therefore, the pH can be calculated as follows:

pOH = -log10 (OH- concentration)
= -log10 (0.06 moles / (50 mL x 1 L / 1000 mL))
= -log10 (0.06 moles / 0.05 L)
= -log10 (1.2 M)

Finally, pH = 14 - pOH
= 14 - (-log10 (1.2 M))

Plug in the calculated value of pOH to get the pH of the solution.