Determine the pH of the following solutions:
a) A 250.0 mL solution that contains 0.125 M benzoic acid and 0.147 M sodium benzoate.
b) The solution in part (a) after 30.0 mL of 0.510 M hydrochloric acid has been added.
a.
Ka = 6.3 x 10^-5
pKa = 4.20
pH = pKa+log(base/acid)
pH = 4.20 + 0.0704
pH = 4.27
b.
HA(aq) + OH-(aq) <----> A-(aq)+H2O(l)
I 0.125M 0.147 - -
C
E
What would i need to do for part b? do i add the hydrochloric acid to the benzoic acid?
I think the easy way to do b is to convert everything to millimols and use that to construct an ICE table. I think this way is easier than converting everything to final M.
mmols HBz = mL x M = 250 x 0.125 = 31.25
mmols Bz^ = 250 x 0.147 = 36.75
mmols HCl added = 15.3
.........Bz^- + H^+ ==> HBz
I.....36.75.....0.......31.25
add............15.3............
C....-15.3....-15.3.....+15.3
E.....21.45.....0.......46.55
Substitute into the HH equation. I came up with 3.86
Yes, in part b, you would need to add the hydrochloric acid to the solution containing benzoic acid and sodium benzoate.
To determine the pH after the addition of hydrochloric acid, you would need to consider the reaction that occurs between hydrochloric acid and benzoic acid. The reaction is as follows:
HA(aq) + HCl(aq) → H2O(l) + ACl(aq)
Since HCl is a strong acid, it completely dissociates in water, so the concentration of HCl in the reaction is equal to the volume of HCl added (30.0 mL) times its concentration (0.510 M).
Using the initial concentrations of benzoic acid (0.125 M), sodium benzoate (0.147 M), and the concentration of HCl, you can set up an ICE table to determine the new equilibrium concentrations.
The table would look as follows:
HA(aq) + OH-(aq) ←→ A-(aq) + H2O(l)
I 0.125M 0.147M - - (Initial concentrations)
C - - x x (Change in concentration)
E 0.125M-x 0.147M-x x x (Equilibrium concentrations)
Since the reaction between HA and OH- is in equilibrium, the concentration of OH- at equilibrium is equal to the change in concentration x.
After determining the new equilibrium concentrations, you can use the Henderson-Hasselbalch equation to calculate the new pH:
pH = pKa + log(A-/HA)
Inserting the appropriate values into the equation will allow you to find the pH of the solution after the addition of hydrochloric acid.
To determine the pH after adding hydrochloric acid in part (b), you need to understand the reaction that occurs between hydrochloric acid (HCl) and benzoic acid (HA). Hydrochloric acid is a strong acid, and it completely dissociates in water to produce H+ ions.
The reaction between HCl and benzoic acid can be represented as follows:
HA(aq) + HCl(aq) --> H3O+(aq) + A-(aq)
Here, HA represents benzoic acid, and A- represents sodium benzoate, which is the conjugate base of benzoic acid.
Since HCl is a strong acid, it completely dissociates in water and provides additional H+ ions. Therefore, the concentration of H3O+ ions increases in the solution.
To determine the resulting pH, you can use the Henderson-Hasselbalch equation again:
pH = pKa + log([base]/[acid])
But this time, you will need to calculate the new concentrations of the benzoic acid and sodium benzoate after the addition of hydrochloric acid.
Let's go through the calculation step by step:
1. Calculate the moles of hydrochloric acid added:
moles HCl = concentration (M) x volume (L)
moles HCl = 0.510 M x 0.0300 L = 0.0153 mol
2. Determine the reaction stoichiometry:
From the balanced equation, we see that 1 mole of HCl reacts with 1 mole of benzoic acid (HA).
3. Calculate the change in moles of benzoic acid:
change in moles HA = moles HCl = 0.0153 mol
4. Update the concentrations of benzoic acid and sodium benzoate:
new concentration of HA = initial concentration of HA - change in moles HA
new concentration of HA = 0.125 M - 0.0153 mol / (0.250 L + 0.0300 L)
new concentration of HA = 0.0917 M
new concentration of A- = initial concentration of A- + change in moles HA
new concentration of A- = 0.147 M + 0.0153 mol / (0.250 L + 0.0300 L)
new concentration of A- = 0.111 M
5. Calculate the new pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])
pH = 4.20 + log(0.111 M/0.0917 M)
pH ≈ 4.32
So, after adding 30.0 mL of 0.510 M hydrochloric acid to the solution in part (a), the pH would be approximately 4.32.