Question

The pH of buffer solutions.

Weight of NaC2H3O2*3H2O =3.47g

pH of original buffer soution = 4.53

pH of buffer + HCl = 3.94

pH of buffer + NaOH = 5.00

Additional details: approximately 8.8 mL of 3M acetic acitd was added to the 3.47g of NaC2H3O2*3H2O. And approximately 56.0mL of distilled water. (This makes the buffer solution) This solution was then split in half. To one half 1 mL of 3M HCl was added. To the second half 1mL of 3M Na OH was added.

Calculate the pH of buffer.

Calculate the pH of buffer + HCl

Calculate the pH of buffer + NaOH

Please show all calculations and work.

please help me. I confused

Answer 1

Mass of NaC2H3O2*3H2O = 3.47 g

Molar mass of NaC2H3O2*3H2O = 136.07 g/mol

Moles of NaC2H3O2*3H2O = 3.47 g / 136.07 g/mol

= 0.0255 moles

Molarity of acitic acid = 3 M

Volume of acetic acid = 8.8 mL = 0.0088 L

Moles of acetic acid = 3.0 * 0.0088

= 0.0264 moles

Total volume of buffer = 56.0 + 8.8

= 64.8 mL = 0.0648 L

[HA] = 0.0264 / 0.0648

= 0.407 M

[A-] = 0.0255 / 0.0648

= 0.393 M

pKa of acetic acid = 4.76

pH = pKa + log [A-] / [HA]

= 4.76 + log(0.393 / 0.407)

= 4.76 - 0.0152

= 4.744

When this solution is divided in two halves.

Volume of each half = 32.4 mL = 0.0324 L

Volume of HCl added = 1 mL = 0.001 L

Molarity of HCl = 3 M

Moles of HCl = 3 * 0.001

= 0.003 M

Total moles of H+ = 0.0264 + 0.003

= 0.0294 moles

Total volume = 32.4 + 1.0 = 33.4 mL

= 0.0334 L

[A-] = 0.0255 / 0.0334

= 0.763 M

[HA] = 0.0294 / 0.0334

= 0.880 M

pH = 4.76 + log (0.763 / 0.880)

= 4.76 - 0.062

= 4.698

When 1mL of 3M Na OH was added to other half.

Moles of OH- = 0.003

Total moles of A- = 0.0255 + 0.003

= 0.0285

[A-] = 0.0285 / 0.0334

= 0.853 M

[HA] = 0.0264 / 0.0334

= 0.790 M

pH = 4.76 + log (0.853 / 0.790)

= 4.76 + 0.033

= 4.76