Consider the titration of 41.2ml of 0.230 M HF with 0.220 M NaOH . Calculate the ph after the addition of 11.9ml of sodium hydroxide.(HF: Ka = 3.50 x 10-4)
Please, I need step by step on on how to solve this type of problem.
Thank you very much!
The secret to these problems is to know where you are on the titration curve; i.e., beginning (0 mL), equilvalence point, before the eq pt or after the eq pt.
You can calculate the eq pt easily.
mL acid x M acid = mL base x M base (that is always true for MONOPROTIC acids titrated with monobasic bases).
41.2 x 0.230 = mL base x 0.220
mL base = estimated 43 mL
Therefore, you know 11.9 mL is between the beginning and the eq. pt and those pHs are determined by the Henderson-Hasselbalch equation.
pH = pKa + log(base)/(acid)
where (base) is the (F^-) and (acid) is the (HF).
mols HF initially = M x L = 0.230 x 0.0412 = 0.00948
mol NaOH added = M x L = 0.220 x 0.0119 = 0.00262
...........HF + NaOH ==> NaF + H2O
I.....0.00948....0........0......0
added..........0.00262..........
C...-0.00262..-0.00262...0.00262......
E.....0.00686...0........0.00262
Ka = 3.5E-4; pKa = -log Ka = 3.46
You can use a shortcut and use mols instead of concn which makes it
pH = 3.46 + log((0.00262/0.00686) = ?
Just a quick note to say that I rounded my numbers above so you should go through the entire calculation and change numbers that I rounded more or less than I should have done. My answer is close to the correct value but it may not be exact.
To solve this problem, you will need to consider the reaction between HF (hydrofluoric acid) and NaOH (sodium hydroxide). This is an acid-base titration problem. Here are the step-by-step instructions to solve it:
Step 1: Write the balanced chemical equation for the reaction between HF and NaOH. The reaction will produce water (H2O) and a salt (NaF).
HF + NaOH -> H2O + NaF
Step 2: Determine the initial moles of HF before the titration. To do this, you need to use the given volume (41.2 mL) and concentration (0.230 M) of HF.
Moles of HF = Volume (in liters) x Concentration
= 41.2 mL / 1000 mL/L x 0.230 M
= 0.009496 moles
Step 3: Determine the initial moles of NaOH before the titration. To do this, you need to use the given volume (0 mL initially) and concentration (0.220 M) of NaOH.
Moles of NaOH = Volume (in liters) x Concentration
= 0 mL / 1000 mL/L x 0.220 M
= 0 moles
Step 4: Determine the number of moles of HF that reacted with the NaOH using stoichiometry. Since the balanced equation is 1:1, the moles of HF reacted will be the same as the moles of NaOH added.
Moles of HF reacted = Moles of NaOH added
= 11.9 mL / 1000 mL/L x 0.220 M
= 0.002618 moles
Step 5: Calculate the moles of HF remaining after the reaction. Subtract the moles of HF reacted from the initial moles of HF.
Moles of HF remaining = Initial moles of HF - Moles of HF reacted
= 0.009496 moles - 0.002618 moles
= 0.006878 moles
Step 6: Calculate the volume of the final solution. Since the volume of the HF solution was 41.2 mL and 11.9 mL of NaOH were added, the total volume of the solution at the end is (41.2 mL + 11.9 mL) = 53.1 mL.
Step 7: Calculate the final concentration of HF in the solution.
Final concentration of HF = Moles of HF remaining / Volume (in liters) of the final solution
= 0.006878 moles / (53.1 mL / 1000 mL/L)
= 0.1295 M
Step 8: Calculate the pOH of the final solution using the concentration of HF (0.1295 M) and the Ka value (3.50 x 10^-4) given for HF.
pOH = -log10[Ka x Concentration of HF]
= -log10[(3.50 x 10^-4) x 0.1295]
Step 9: Calculate the pH of the final solution using the pOH.
pH = 14 - pOH
Following these steps should help you calculate the pH after adding 11.9 mL of NaOH to the HF solution.