Question

Starting with 0.25 moles of sodium phosphate (Na3PO4) in 1.00 L volumetric flask, determine the pH of the resulting solution upon addition of the amounts of hydrochloric acid shown in the table below. Each of the solutions were diluted to a final volume of 1.00 L.

Moles of HCl	pH
0.00
0.15
0.25
0.75
1.00

Answer 1

1) for 0.00 mols of HCL

mole of Na3PO4 = 0.25

mole of HCl = 0

After these react, 0 H+ is left

Con of H+ = no of mole of H+ / total volume of solution (1000 ml)

= 0/100 = 0

pH =- log ( H+)

=- log (0) = 0

2) 0.15 mol of HCL

mole of Na3PO4 = 0.25

mole of HCl = 0.15

After these react, (0.25-0.15=0.1) H+ is left

Con of H+ = no of mole of H+ / total volume of solution (1000 ml)

= 0.1/1000 = 0.0001

pH = -log ( H+)

=- log (0.0001) = 4

2) 0.25 mol of HCL

mole of Na3PO4 = 0.25

mole of HCl = 0.25

After these react, 0 H+ is left

Con of H+ = no of mole of H+ / total volume of solution (1000 ml)

= 0/1000 = 0

pH = -log ( H+)

=- log (0.000) = 0

3) 0.75 mol of HCL

mole of Na3PO4 = 0.25

mole of HCl = 0.75

After these react, (0.75-0.25=0.5 ) H+ is left

Con of H+ = no of mole of H+ / total volume of solution (1000 ml)

= 0.5/1000 = 0.0005

pH = -log ( H+)

=- log (0.0005) = 3.30

1) 1 mol of HCL

mole of Na3PO4 = 0.25

mole of HCl = 0.1

After these react, (1-0.25=0.75) H+ is left

Con of H+ = no of mole of H+ / total volume of solution (1000 ml)

=0.75/1000 = 0.00075

pH = -log ( H+)

=- log (0.00075) = 3.12