(salt )(acid )

(HPO 4−2)

¿¿

(HPO 4−2)

¿¿

(HPO 4−2)

¿¿0.6166n/L1.0000n/L

H3PO4 + NaOH NaH2PO4 + H2O

NaH2PO4 + NaOH Na2HPO4 + H2O

pH = pKa + log

7.0 = 7.21 + log

7.0 – 7.21 = log

antilog (-0.21) = 0.6166

ratio of salt and acid

=

Theoretical number of moles

0.6166 n HPO 4−2

1.0000 n H 2PO4−¿¿

1.6166 n

0.025 n 0.025 n

0.010 n

pKa 2.12

pKa 7.21

+

0.010 n

0.015 n

Actual number of moles of phosphate buffer250mL×

1L1000mL

=0.250 L

0.1nL×0.250 L=0.025n

Actual number of moles of HPO 4−2

0.6166n1.6166n

=n HPO4

−2

0.025n

HPO 4−2 = 0.010 n

Actual number of moles of H 2PO4−¿¿

1.0000n1.6166n

=n H 2PO 4

−¿

0.025n¿

H 2PO4−¿¿ = 0.015 n

Amount needed if materials available are 15 M H3PO4 and 6 M NaOH

mL of H3PO4 needed0.025n×

1 L15n

×1000mL

1L=1.67mLH 3PO 4

mL of 6 M NaOH

0.025 n + 0.010 n = 0.035 n0.0 3 5n×

1L6n×

1000mL1L

=5.83mLNaOH