Calculate the pH during the titration of 20.00 mL of 0.1000 M CH3CH2CH2COOH with 0.1000 M NaOH after 26.5 mL of the base have been added. Ka of butanoic acid = 1.54 x 10-5.
To calculate the pH during the titration of butanoic acid (CH3CH2CH2COOH) with sodium hydroxide (NaOH), we need to determine the amount of acid and base present after 26.5 mL NaOH has been added.
1. First, calculate the number of moles of butanoic acid initially present:
moles of CH3CH2CH2COOH = (volume in liters) x (concentration in M)
moles of CH3CH2CH2COOH = (20.00 mL / 1000 mL/L) x (0.1000 M)
moles of CH3CH2CH2COOH = 0.00200 moles
2. Next, determine the number of moles of NaOH added:
moles of NaOH = (volume in liters) x (concentration in M)
moles of NaOH = (26.5 mL / 1000 mL/L) x (0.1000 M)
moles of NaOH = 0.00265 moles
3. Since both the acid and base have a 1:1 stoichiometric ratio, the number of moles of butanoic acid remaining can be calculated:
moles of CH3CH2CH2COOH remaining = moles of CH3CH2CH2COOH initially - moles of NaOH added
moles of CH3CH2CH2COOH remaining = 0.00200 moles - 0.00265 moles
moles of CH3CH2CH2COOH remaining = -0.00065 moles
Note: A negative value indicates that all of the butanoic acid has been neutralized.
4. Calculate the volume of NaOH required to reach the equivalence point when all of the acid has been neutralized:
volume of NaOH at equivalence point = (moles of CH3CH2CH2COOH initially) / (concentration of NaOH)
volume of NaOH at equivalence point = (0.00200 moles) / (0.1000 M)
volume of NaOH at equivalence point = 0.0200 L or 20.0 mL
5. Calculate the volume of NaOH added after the equivalence point:
excess volume of NaOH = (volume of NaOH added) - (volume of NaOH at equivalence point)
excess volume of NaOH = 26.5 mL - 20.0 mL
excess volume of NaOH = 6.5 mL
6. Using the excess volume of NaOH, calculate the number of moles of excess NaOH:
moles of excess NaOH = (volume in liters) x (concentration in M)
moles of excess NaOH = (6.5 mL / 1000 mL/L) x (0.1000 M)
moles of excess NaOH = 0.00065 moles
7. Finally, calculate the concentration of the resulting solution:
concentration of CH3CH2CH2COOH = (moles of CH3CH2CH2COOH remaining) / (volume of resulting solution)
concentration of CH3CH2CH2COOH = (-0.00065 moles) / [(20.00 mL + 6.5 mL) / 1000 mL/L]
concentration of CH3CH2CH2COOH = -0.0200 M
The negative concentration value suggests that the solution is basic, so we need to determine the concentration of the conjugate base (CH3CH2CH2COO-) resulting from the reaction between the acid and base.
8. Calculate the concentration of CH3CH2CH2COO-:
concentration of CH3CH2CH2COO- = (moles of NaOH added) / (volume of resulting solution)
concentration of CH3CH2CH2COO- = 0.00265 moles / [(20.00 mL + 6.5 mL) / 1000 mL/L]
concentration of CH3CH2CH2COO- = 0.0678 M
Now, we can calculate the pH of the resulting solution using the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
pKa = -log(Ka) = -log(1.54 x 10^-5) = 4.81 (rounded to two decimal places)
[A-] = concentration of the conjugate base = 0.0678 M (approximated)
[HA] = concentration of the acid = -0.0200 M (approximated, negative concentration)
pH = 4.81 + log(0.0678 M / -0.0200 M)
Note: The negative concentration of the acid causes the log term to be undefined, indicating that the pH cannot be determined in this case.
The secret to these titration problems is to know where you are on the titration curve.
Let's call the acid HBu.
HBu + NaOH ==> NaBu + H2O
mmoles HBu = M x L = 20.00 mL x 0.1 = 2 mmoles.
mmoles NaOH = M x L = 26.5 mL x 0.1 = 2.65.
It should be obvious that all of the HBu acid has been neutralized, you have passed the equivalence point and the pH is determined by the excess OH^- in the solution.
So how much past? 2.65 mmoles - 2.00 mmoles = 0.65 mmoles and the concn OH^- = 0.65 mmoles/(20 + 26.5)mL = ??
Then convert to pOH and to pH.