A 50.0 mL sample of 0.23 M propanoic acid, CH3CH2COOH, a weak monoprotic acid, is titrated with 0.14 M KOH. Ka of CH3CH2COOH = 1.4 multiplied by 10-5.
What is the pH at equivalence point?
The pH will be determined by the hydrolysis of the salt, sodium propionate.
If we abbreviate propanoic acid as HPr, then the titration is
HPr + NaOH ==> NaPr + H2O
The salt is NaPr and it's the Pr^- that is hydrolyzed.
Pr^- + HOH ==> HPr + OH^-
Set up an ICE chart and substitute into the below.
Kb = (Kw/Ka) = (HPr)(OH^-)/(Pr^-)
Kw = 1E-14
Ka = the value for HPr
(HPr) = (OH^-) = x
(Pr^-) = (moles/L pr millimoles/mL) which you will need to calculate.
You had 50 mL x 0.23 M HPr = ?millimoles and that will require mmoles/0.14 = about 80 mL NaOH to reach the equivalence point (you need to do the 80 more accurately). Then total volume is 80 or so + 50 = ??.
To find the pH at the equivalence point, we need to determine the concentration of the resulting salt and then calculate the concentration of hydroxide ions (OH-) produced. Finally, we can use the concentration of hydroxide ions to find the pH.
At the equivalence point, the moles of the acid will be equal to the moles of the base. This means that the moles of propanoic acid (CH3CH2COOH) will react completely with the moles of potassium hydroxide (KOH).
First, let's find the moles of propanoic acid:
Moles of propanoic acid = Concentration of propanoic acid × Volume of propanoic acid in liters
Moles of propanoic acid = 0.23 M × (50.0 mL / 1000 mL) [Converting mL to liters]
Moles of propanoic acid = 0.0115 moles
Since propanoic acid is a monoprotic acid, it will react with an equal number of moles of hydroxide ions (OH-). Therefore, the concentration of hydroxide ions will be equal to the concentration of potassium hydroxide (KOH) used in titration.
Next, let's find the concentration of hydroxide ions:
Concentration of hydroxide ions = Concentration of KOH
Concentration of hydroxide ions = 0.14 M
Now, let's calculate the pOH at the equivalence point:
pOH = -log10[OH-]
pOH = -log10(0.14)
Using the equation pOH + pH = 14, we can find the pH:
pH = 14 - pOH
Substituting the value for pOH:
pH = 14 - (-log10(0.14))
Finally, we can calculate the pH by substituting the values into the equation and evaluating:
pH ≈ 11.85
Therefore, the pH at the equivalence point is approximately 11.85.