Find the pH during the titration of 20.00 mL of 0.1000 M butanoic acid, CH3CH2CH2COOH (Ka = 1.54e-5), with 0.1000 M NaOH solution after the following additions of titrant: 0 mL, 14.00 mL, 19.95 mL, 20.00 mL, 20.05 mL, 24.00 mL.
To find the pH during the titration of butanoic acid (CH3CH2CH2COOH) with NaOH, we need to analyze the effect of added NaOH on the acid-base equilibrium. Let's tackle each addition step by step:
1. 0 mL NaOH:
At this point, only butanoic acid is present in the solution. To find its pH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
Here, [A-] represents the concentration of the conjugate base of butanoic acid, and [HA] represents the concentration of butanoic acid.
Since no NaOH has been added yet, [A-] = 0, and [HA] = 0.1000 M (initial concentration). So, the pH can be calculated using the known pKa value of 1.54e-5.
2. 14.00 mL NaOH:
At this point, a fraction of butanoic acid has been neutralized by the added NaOH. The number of moles of NaOH added can be calculated using the volume and concentration of NaOH.
With the number of moles of NaOH, we can determine the number of moles of butanoic acid that reacted.
From the stoichiometry of the reaction, we can determine how the concentration of butanoic acid and its conjugate base have changed.
Again, we can use the Henderson-Hasselbalch equation to calculate the pH using the updated concentrations.
3. 19.95 mL NaOH and 20.05 mL NaOH:
These volumes fall close to the equivalence point, where the moles of NaOH added are stoichiometrically equivalent to the moles of butanoic acid initially present. This means that the concentration of butanoic acid and its conjugate base will change significantly.
At the equivalence point, the pH will depend on factors like the dissociation constant (Ka) and the volumes of the acid and base solutions used. We need to perform more calculations to find the pH precisely using the updated concentrations.
4. 24.00 mL NaOH:
At this point, excess NaOH has been added. The reaction between butanoic acid and NaOH is no longer stoichiometric, and hence the pH will be dependent on the concentration of NaOH remaining and the excess volume added. Again, calculations are needed to determine the pH.
To summarize, the pH at each titration step requires careful consideration of the acid-base equilibrium and relevant calculations using the Henderson-Hasselbalch equation, stoichiometry, and the known concentrations and volumes of the acid and base solutions.
The secret to doing this is to recognize what you have at each titration point. If we let HB stand for butanoic acid, then
the equation is HB + NaOH ==> NaB + H2O
a. at the beginning you have just HB. Set up an ICE chart
............HB ==> H^+ + B^-
initial....0.1.......0....0
change.....-x.........x....x
equil.....0.1-x.......x.....x
Substitute into the Ka expression, solve for H^+, and convert to pH.
Next determine the equivalence point from mL HB x MHB = mLNaOH x MNaOH
The pH at the equivalence point is determined by the concn of the salt produced. It is hydrolyzed. Set up an ICE chart for that.
.........B^- + HOH ==> HB + OH^-
initial..C.............0.....0
change...-x............x......x
equil....C-x............x.....x
Kb = (Kw/Ka) = (HB)(OH^-)/(C)
Kw you know. Ka you know.
C is determined from the initial equation at the top of the page and is moles/L = M. Solve for HB = x= OH^-, convert to pOH, then to pH.
All points leading up to the equivalence point are buffered solutions, partly HB an partly NaB and can be solved with the Henderson-Hasselbalch equation.
All points after the equivalence point are due to the excess HCl. Calculate excess HCl (don't forget this is diluted by the volumes) and convert to pH.
Post your work if you get stuck.