What is the ph of a soft drink in which the major buffer ingredients are 6.6 g of nah2po4 and 8.0 g of na2hpo4 per 355 ml of solution?

The Relative Formula Mass of NaH2PO4 is 120 g/mol

Therefore, the number of moles = 6.6/120

= 0.055 moles of NaH2PO4 which is also equal to the number of moles of H2PO4.

[H2PO4-] = Number of moles oof H2PO4-/Volume of the solution in L

= 0.055 / (355 * 10^-3)

= 0.155 M

Na2HPO4 undergoes complete dissociation as follows;

Na2HPO4 (aq) = 2Na + (aq) + HPO4^2 - (aq)

1 mole of Na2HPO4 = 142 g/mol

Therefore; number of moles = 8.0/142

= 0.0563 moles

[HPO4 ^-2] is given by no of moles HPO4^2 - / volume of the solution in L

= 0.0563 / (355*10^-3)

= 0.1586 M

Both H2PO4^2 - and HPO4^2 - are weak acids the undergoes partial dissociation

Ka of H2PO4 - = 6.20 * 10^-8

[H+] = Ka * ([H2PO4-]/[HPO4 (2-) ]

= (6.20 * 10^-8) * (0.155/0.1586)

= 6.059 * 10^-8 M

pH = - log[H+]

= - log (6.059*10^-8)

= 7.218