The pKa of butanoic acid is 4.87. A student mixes 120 mL of 0.45 M butanoic acid
with 50 mL of 0.3 M NaOH and adds water to give a final volume of 1 L. What is the
pH of the final solution? Please show you work.
To find the pH of the final solution, we need to determine the moles of butanoic acid and NaOH present in the mixture and then calculate the concentrations of the resulting ions. Finally, we can use the equilibrium constant expression and the Henderson-Hasselbalch equation to find the pH.
Step 1: Calculate the moles of butanoic acid:
Molarity of butanoic acid = 0.45 M
Volume of butanoic acid = 120 mL = 0.12 L
Moles of butanoic acid = Molarity x Volume
= 0.45 M x 0.12 L
= 0.054 moles
Step 2: Calculate the moles of NaOH:
Molarity of NaOH = 0.3 M
Volume of NaOH = 50 mL = 0.05 L
Moles of NaOH = Molarity x Volume
= 0.3 M x 0.05 L
= 0.015 moles
Step 3: Determine the concentrations of the resulting ions:
The reaction between butanoic acid and NaOH is a neutralization reaction. It produces water and sodium butanoate.
According to the balanced equation:
Butanoic acid + NaOH → Sodium butanoate + Water
The stoichiometry of the reaction is 1:1. This means that for every mole of butanoic acid that reacts, one mole of sodium butanoate is produced.
Since the moles of butanoic acid and NaOH are the same, the resulting moles of sodium butanoate will be equal to the moles of reactant that were consumed.
Therefore, the moles of sodium butanoate = 0.054 moles.
Now, let's calculate the concentrations:
Concentration of sodium butanoate = moles of sodium butanoate / final volume
= 0.054 moles / 1 L
= 0.054 M (this is also the concentration of butanoic acid)
The concentration of water in the final solution is the same as the final volume, which is 1 L.
Step 4: Calculate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log (concentration of sodium butanoate / concentration of butanoic acid)
pKa of butanoic acid = 4.87
Concentration of sodium butanoate = 0.054 M
pH = 4.87 + log (0.054 M / 0.054 M)
= 4.87 + log (1)
= 4.87 + 0
= 4.87
Therefore, the pH of the final solution is 4.87.
To calculate the pH of the final solution, we need to determine the concentration of the acidic and basic components and then apply the appropriate equations.
Step 1: Calculate the number of moles of butanoic acid (CH3CH2CH2COOH) and NaOH (sodium hydroxide):
Moles of butanoic acid = volume (L) × concentration (M) = 0.120 L × 0.45 M
Moles of NaOH = volume (L) × concentration (M) = 0.050 L × 0.3 M
Step 2: Determine the limiting reagent:
The limiting reagent is the component that is completely consumed in the reaction. In this case, butanoic acid and NaOH react in a 1:1 ratio (according to their balanced equation). To determine the limiting reagent, compare the moles of each component. The component with fewer moles will be the limiting reagent.
Moles of butanoic acid = 0.120 L × 0.45 M = 0.054 moles
Moles of NaOH = 0.050 L × 0.3 M = 0.015 moles
Since NaOH has fewer moles, it is the limiting reagent.
Step 3: Determine the remaining moles of the excess reactant:
The remaining moles of butanoic acid will be:
Moles of butanoic acid = initial moles - consumed moles
Moles of butanoic acid = 0.054 moles - 0.015 moles = 0.039 moles
Step 4: Calculate the concentration of butanoic acid in the final solution:
Concentration (in M) = moles / volume (in L) = 0.039 moles / 1 L = 0.039 M
Step 5: Determine the pH of the solution using the Henderson-Hasselbalch equation:
pH = pKa + log10([A-]/[HA])
In this case, butanoic acid (CH3CH2CH2COOH) acts as an acid, so [A-] is the concentration of the conjugate base (CH3CH2CH2COO-) and [HA] is the concentration of the acid (CH3CH2CH2COOH).
pH = 4.87 + log10([CH3CH2CH2COO-] / [CH3CH2CH2COOH])
Since the concentration of butanoic acid is 0.039 M, [CH3CH2CH2COOH] = 0.039 M.
To calculate the concentration of the conjugate base, we need to consider the reaction of butanoic acid with NaOH. The balanced equation is:
CH3CH2CH2COOH + NaOH → CH3CH2CH2COO- + H2O
Since both butanoic acid and NaOH react in a 1:1 ratio, the concentration of the conjugate base will also be 0.039 M.
pH = 4.87 + log10(0.039 M / 0.039 M) = 4.87
Therefore, the pH of the final solution is 4.87.