Benzoic acid (C6H5COOH)is a monoprotic weak acid with Ka=6.30*10^-5. What is the pH of a solution of benzoic acid that is 0.559M and has 2.25*10^-2M NaOH added?
By adding NaOH you have formed a buffered solution. Take 1 L soln, so you have
1000 x 0.559 M = 559 millimols PhCOOH.
1000 x 0.0225M = 22.5 mmols NaOH
.......PhCOOH + NaOH ==> PhCOONa + H2O
initial..559................0........0
added...........22.5...................
change....-22.5..-22.5.....22.5
equil....536.5...0..........22.5
Use the Henderson-Hasselbalch equation and solve for pH.
To find the pH of the solution, we need to consider the dissociation of benzoic acid and the neutralization reaction with sodium hydroxide (NaOH).
First, let's write the balanced equation for the dissociation of benzoic acid:
C6H5COOH ⇌ C6H5COO- + H+
Since benzoic acid is a weak acid, we can assume that the concentration of the benzoate ion (C6H5COO-) at equilibrium will be negligible compared to the initial concentration of benzoic acid.
Next, let's write the balanced equation for the neutralization reaction between benzoic acid and sodium hydroxide:
C6H5COOH + NaOH → C6H5COONa + H2O
From the balanced equation, we can see that for every one mol of benzoic acid that reacts, one mol of hydroxide ion (OH-) is consumed, resulting in the formation of water (H2O).
Since the NaOH concentration is given as 2.25*10^-2M, we can assume that the concentration of OH- added to the solution will be 2.25*10^-2M.
Now, let's calculate the moles of benzoic acid and OH- in the solution:
moles of benzoic acid = 0.559M * volume of solution in liters
moles of OH- = 2.25*10^-2M * volume of solution in liters
Assuming the volumes cancel out, we can proceed with the calculations.
Next, we need to determine how much of the benzoic acid has been neutralized by the OH-. Since benzoic acid is a weak acid, it will not fully dissociate, so we need to calculate the amount of benzoic acid that has reacted.
The reaction between benzoic acid and OH- occurs in a 1:1 ratio, meaning that for every mol of OH- that reacts, one mol of benzoic acid will be consumed.
Therefore, the amount of benzoic acid that has reacted is equal to the amount of OH- added to the solution.
Now, let's calculate the remaining concentration of benzoic acid after neutralization:
remaining concentration of benzoic acid = initial concentration - amount reacted
Finally, we can use the concentration of the remaining benzoic acid to calculate the pH of the solution by using the Ka value:
pH = -log([H+])
[H+] = √(Ka * remaining concentration of benzoic acid)
Now we have all the necessary information and calculations to determine the pH of the solution.