A 500 mL buffer solution is 0.1 M benzoic acid and 0.10 M in sodium benzoate and has an initial pH of 4.19. What is the pH of the buffer upon addition of 0.010 mol of NaOH?
To determine the pH of the buffer upon addition of NaOH, we need to consider the reaction between benzoic acid (a weak acid) and sodium hydroxide (a strong base). The reaction can be represented as follows:
C6H5COOH (benzoic acid) + NaOH (sodium hydroxide) → C6H5COONa (sodium benzoate) + H2O
Given the concentrations of benzoic acid and sodium benzoate, we can calculate the initial moles of benzoic acid in the buffer solution.
Moles of benzoic acid = volume of buffer solution (L) × concentration of benzoic acid (M)
= 0.5 L × 0.1 M
= 0.05 mol
Since the ratio of benzoic acid to sodium benzoate in the buffer solution is 1:1, the initial moles of sodium benzoate are also 0.05 mol.
After adding 0.010 mol of NaOH, the total moles of sodium benzoate in the buffer solution will increase to 0.05 + 0.010 = 0.06 mol.
To calculate the pH, we need to determine the concentrations of benzoic acid and sodium benzoate after the addition of NaOH:
Concentration of benzoic acid after the addition of NaOH = moles of benzoic acid / total volume of buffer solution (L)
= 0.05 mol / 0.5 L
= 0.1 M
Concentration of sodium benzoate after the addition of NaOH = moles of sodium benzoate / total volume of buffer solution (L)
= 0.06 mol / 0.5 L
= 0.12 M
To calculate the pH of the buffer solution after the addition of NaOH, we can use the Henderson-Hasselbalch equation:
pH = pKa + log10([A-]/[HA])
Where:
- pH is the desired value we are trying to find.
- pKa is the dissociation constant of the weak acid (benzoic acid). In this case, the pKa of benzoic acid is known to be about 4.2.
- [A-] is the concentration of the conjugate base (sodium benzoate), which is 0.12 M.
- [HA] is the concentration of the weak acid (benzoic acid), which is 0.1 M.
Now we can substitute these values into the equation:
pH = 4.2 + log10(0.12/0.1)
= 4.2 + log10(1.2)
≈ 4.2 + 0.08
≈ 4.28
Therefore, the pH of the buffer solution after the addition of 0.010 mol of NaOH is approximately 4.28.
To find the pH of the buffer upon addition of NaOH, we need to calculate the new concentrations of benzoic acid and sodium benzoate after the reaction with NaOH.
The balanced chemical equation for the reaction between sodium hydroxide (NaOH) and benzoic acid (C6H5COOH) is:
C6H5COOH + NaOH -> C6H5COONa + H2O
From the equation, we can see that one mole of NaOH reacts with one mole of benzoic acid to form one mole of sodium benzoate (C6H5COONa) and one mole of water (H2O).
Given that 0.010 mol of NaOH is added and that benzoic acid and sodium benzoate are in equal molar concentrations initially, we can assume that 0.010 mol of benzoic acid will be consumed. This results in a decrease of 0.010 M for benzoic acid and an increase of 0.010 M for sodium benzoate.
Since the initial concentrations are 0.1 M for benzoic acid and 0.10 M for sodium benzoate, the new concentrations after the addition of NaOH will be:
Benzoic acid: 0.1 M - 0.010 M = 0.090 M
Sodium benzoate: 0.10 M + 0.010 M = 0.110 M
To calculate the pH of the buffer, we can use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Where pKa is the acid dissociation constant of benzoic acid and [A-] / [HA] is the ratio of the concentration of sodium benzoate to benzoic acid.
The pKa of benzoic acid is approximately 4.20.
Plugging in the values, we get:
pH = 4.20 + log (0.110 / 0.090)
Calculating the log ratio yields:
pH = 4.20 + log (1.222)
Using a calculator, we find:
pH ≈ 4.76
Therefore, the pH of the buffer upon addition of 0.010 mol of NaOH is approximately 4.76.
Let's call benzoic acid HB and sodium benzoate is NaB. NaB is the base; HB is the acid.
.......HB + NaOH ==> NaB + H2O
I......0.1....0.......0.1......
add.........0.01..............
C....-0.01..-0.01......+0.01
E.....0.09.....0.......0.11
Substitute the E line into the Henderson-Hasselbalch equation and solve for pH.