Do parts a, b and c a. calculate the ph of 100 ml of an aqueous solution of 0.05 m benzoic acid and 0.025 m benzoate ion. the ka of benzoic acid is 6.3 x 10-5 at 25 ℃. b. calculate the new ph of the solution in part a if you add 10.0 ml of 0.100 m hcl solution. c. calculate the new ph of the solution in part a if you add 15.0 ml of 0.050 m naoh solution.

- (a)The pH of the buffer solution is 3.90.

(b) the pH of the solution after addition of HCl would be 3.60.

(c) the pH of the buffer solution after addition of NaOH is 4.32.

Solution:- (a) It is a buffer solution so the pH could easily be calculated using Handerson equation:

pKa can be calculated from given Ka value as:

pKa = 4.20

let's plug in the values in the Handerson equation:

pH = 4.20 - 0.30

pH = 3.90

The pH of the buffer solution is 3.90.

(b) Let's say the acid is represented by HA and the base is represented by .

Original mili moles of HA from part a = 0.05(100) = 5

original mili moles of from part a = 0.025(100) = 2.5

mili moles of HCl that is added = 0.100(10.0) = 1

This HCl reacts with the base present in the buffer to make HA as:

Total mili moles of HA after addition of HCl = 5+1 = 6

mili moles of base after addition of HCl = 2.5-1 = 1.5

Let's plug in the values in the Handerson equation again. Here, we could use the mili moles to calculate the pH. The answer remains same even if we use the concentrations also as the final volume is same both for acid and base.

pH = 4.20 - 0.60

pH = 3.60

So, the pH of the solution after addition of HCl would be 3.60.

(c) mili moles of NaOH or added to the original buffer = 0.05(15.0) = 0.75

This reacts with HA to form as:

mili moles of HA after addition of NaOH = 5-0.75 = 4.25

mili moles of after addition of NaOH = 2.5+0.75 = 3.25

Let's plug in the values again in Handerson equation:

pH = 4.20 - 0.12

pH = 4.32

So, the pH of the buffer solution after addition of NaOH is 4.32.

they both have a y in them