Calculate the pH of the 1L buffer composed of 500 mL 0.60 M acetic acid plus 500 mL of 0.60 M sodium acetate, after 0.010 mol of NaOH is added (Ka HC2H3O2 = 1.75 x 10-5). Report your answer to the hundredths place.

pH = 4.79

Explanation:

The pH of the acetic buffer can be determined using H-H equation:

pH = pKa + log [A⁻] / [HA]

Where pKa is -logKa = 4.76

pH = 4.76 + log [sodium Acetate] / [Acetic Acid]

Where [] can be taken as moles of each specie.

Thus, to find pH of the buffer we need to calculate molesof acetic acid and sodium acetate.

Initial moles:

Initial moles of acetic acid and sodium acetate are:

500mL = 0.500L ₓ (0.60moles / L) = 0.30 moles of both acetic acid and sodium acetate

Moles after reaction:

Now, 0.010 moles of NaOH are added to the buffer reacting with acetic acid, CH₃COOH, producing more acetate ion, as follows:

NaOH + CH₃COOH → CH₃COO⁻ + H₂O

That means after reaction moles of both species are:

Acetic acid: 0.30mol - 0.010mol (Moles that react) = 0.29 moles

Acetate: 0.30mol + 0.010mol (Moles produced) = 0.31 moles

Replacing in H-H equation:

pH = 4.76 + log [0.31] / [0.29]

answer; economic order quantity;

thats how its supposed to be