A buffer was prepared by dissolving 0.40 mol of benzoic acid (Ka=6.3*10^-5) and 0.50 mol of sodium benzoate in sufficient pure water to form a 1.00 L solution. To this solution was added 30.00 mL of 2.00 M HCl solution. What was the pH of the new solution?
C6H5COOH (aq) + H20 (l) <--------> H3O+ (aq) + C6H5COO- (aq)
The Henderson-Hasselbalch equation is pH = pKa + log [(base)/(acid)]. Technically, those are concentrations for base and acied; however, I will take a shortcut and use mols since the volume cancels. Your prof may not like the shortcut so beware. Also, I will let HB stand for benzoic acid and B^- is benzoate.
mols HCl added = M x L = 0.06
...................B^- + H^+ ==> HB
I................0.5......0.............0.4
add.....................0.06..............
C...........-0.06..-0.06...........+0.06
E............0.44.......0.............0.46
Plug the E line into the HH equation and solve for the new pH.
Post your work if you get stuck.
To determine the pH of the new solution, we need to calculate the change in concentration of the conjugate base and conjugate acid after the addition of HCl and then use the Henderson-Hasselbalch equation to find the new pH.
Step 1: Calculate the concentrations of the species present in the buffer before the addition of HCl.
We know that we have 0.40 mol of benzoic acid (C6H5COOH) and 0.50 mol of sodium benzoate (C6H5COONa) in 1.00 L of solution.
The concentration of benzoic acid (C6H5COOH) can be calculated by dividing the number of moles (0.40 mol) by the volume of the solution (1.00 L):
Concentration of C6H5COOH = 0.40 mol / 1.00 L = 0.40 M
Similarly, the concentration of sodium benzoate (C6H5COONa) can be calculated by dividing the number of moles (0.50 mol) by the volume of the solution (1.00 L):
Concentration of C6H5COONa = 0.50 mol / 1.00 L = 0.50 M
Step 2: Calculate the change in concentration of the conjugate base and conjugate acid after the addition of HCl.
We are adding 30.00 mL (0.030 L) of 2.00 M HCl to the buffer solution.
The number of moles of HCl added can be calculated by multiplying the concentration (2.00 M) by the volume (0.030 L):
Number of moles of HCl added = 2.00 M × 0.030 L = 0.060 mol
Since benzoic acid (C6H5COOH) is a weak acid, it reacts with HCl to form H3O+ (aq) and C6H5COO- (aq). The reaction between C6H5COOH and HCl is a 1:1 ratio, so the number of moles of C6H5COOH that react is also 0.060 mol.
After the reaction, the change in concentration of C6H5COOH is:
Change in concentration of C6H5COOH = initial concentration - moles reacted
= 0.40 M - 0.060 mol / 1.00 L
= 0.34 M
Therefore, the new concentration of C6H5COOH is 0.34 M.
Step 3: Calculate the concentration of the conjugate base, C6H5COO- (aq).
The reaction between C6H5COOH and HCl results in the formation of an equal number of moles of the conjugate base (C6H5COO-) and H3O+. Therefore, the change in concentration of C6H5COO- is also 0.060 mol, but expressed as a concentration, it is:
Change in concentration of C6H5COO- = moles reacted / final volume
= 0.060 mol / 1.00 L
= 0.060 M
Therefore, the new concentration of C6H5COO- is 0.060 M.
Step 4: Use the Henderson-Hasselbalch equation to calculate the pH of the new solution.
The Henderson-Hasselbalch equation is given by pH = pKa + log ([conjugate base] / [conjugate acid]).
The pKa value for benzoic acid (C6H5COOH) is given as 6.3 × 10^-5.
Substituting the known values into the equation:
pH = -log(6.3 × 10^-5) + log (0.060 / 0.34)
Calculating the logarithms and then evaluating the expression gives us the pH of the new solution.