Adding a Strong Acid to a Buffer?

Question

have the solution, but I still dont understand some of the steps.

A beaker with 115mL of an acetic acid buffer with a pH of 5.00 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.1M . A student adds 4.30mL of a 0.490M HCl solution to the beaker. How much will the pH change? The Pk_a of acetic acid is 4.76.

SOLUTION:

pH = pKa log acetate / acid

5.00 = 4.76 log acetate / acid

acetate / acid =10^0.24= 1.74

acetate acid = 0.1

Solve the system

acetate = 0.0365 M

acid = 0.0635 M

*HOW DID THEY FIND THAT IT'S .0365 M & .0635 M???

*CAN SOMEONE SHOW ME HOW TO THE MATH FOR IT?????

moles acetate = .115 L x 0.0365 M = 0.00420 mol

moles acid = .115 L x 0.0635 M = 0.00730 mol

Moles H added = .00430 L x 0.490 M = 0.00211mol

CH3COO- H --> CH3COOH

moles acetate = 0.00420 - 0.00211 = 0.00209 mol

moles acid = 0.00730 0.00211 = 0.00941 mol

*WHY SUBTRACT FOR ACETATE BUT THEN ADD FOR ACID??

Total volume = 115 4.30 = 119.3 mL = 0.1193 L

concentration acetate = 0.00209/0.1193 = 0.0175 M

concentration acid = 0.00941 / 0.1193 = 0.0789 M

pH = 4.76 log 0.0175 / 0.0789 = 4.11

Answer 1

[CH3COO-] / [CH3COOH] = 1.74

[CH3COOH] + [CH3COO-] = 0.1

[CH3COO-] = 0.1 - [CH3COOH]

We put this value in the 1st equation :

0.1 -[CH3COOH] / [CH3COOH] = 1.74

we multuply the left and the right side by [CH3COOH]

0.1 - [CH3COOH] = 1.74 [CH3COOH]

0.1 = 2.74 [CH3COOH]

[CH3COOH] = 0.0365 M

[CH3COO-] + 0.0365 = 0.1

[CH3COO-] = 0.1 - 0.0365 = 0.0635 M

the effect of the added 0.00211 mol of H+ would be to decrease the moles of CH3COO- by 0.00211 and increase the moles of CH3COOH by 0.00211 by the reaction :

CH3COO- + H+ >> CH3COOH