Question

Calculate the change in pH that occurs when 0.00100 mol of gaseous HCl is added to a buffer solution that is prepared by dissolving 4.92 g of sodium acetate ( molar mass= 82.03 g/mol) in 250 mL of 0.150 mol/L of acetic acid solution? Assume no change in volume upon addition of either the sodium acetate or HCl. (Ka for acetic acid = 1.8x10^-5) *please explain!*

Answer 1

The solution is250mL of 0.150M acetic acid solution.

Now 4.92 g of sodium acetate is dissolved in it.

Molarity of sodium acetate = (4.92/82.03g/mol) (1000/250)

= 0.2399 M

The solution now forms a buffer consisting of a weak acid (acetic acid) and its conjugate base(sodium acetate).

Its pH is calculated using Hendersen equation

pH = pka + log [conjugate base]/[acid]

= (5 - log 1.8) + log (0.2399/0.150)

= 4.9439

When 0.0010 moles of Hcl is added, the reaction that takes place is

CH3COONa + HCl -------> CH3COOH + NaCl

0.2399 - 0.150 - initial concentrations

0.001 change

0.2389 0 0.151 - after reaction

New pH of the solution = 4.74 + log (0.2389/0.151)

= 4.9392

Thus the chang e in pH = 4.9439 -4.9392

=0.0046 6