Question

What is the pH of 100 mL of a solution prepared from 0.600 g of sodium phosphate dodecahydrate mixed with 2.25 mL of 0.995 M hydrochloric acid diluted to to 100 mL with water? Hint: The chemist is preparing a buffer. What two species (a conjugate acid-base pair) are present after all the HCL is consumed? *Please show steps to understand completely*

Answer 1

Solution :-

sodium phosphate dodecahydrate =Na3PO4 .12H2O molar mass is 380.1240 g/mol

lets first calculate its moles

moles = mass / molar mass

           = 0.600 g / 380.1240 g/mol

          = 0.001578 mol

moles of HCl = molarity * volume in liter

                        = 0.995 mol per L * 0.00225 L

                       = 0.002239 mol HCl

After the reaction 0.001578 mol HPO4^2- will form

then moles of HCl remain = 0.002239 - 0.001578 = 0.000 66 mol

Therefore it again protonate the HPO4^2- to form the H2PO4^-

so the moles of HPO4^2- remain = 0.001578 - 0.00066 = 0.000918

and moles of H2PO4^- formed = 0.00066 mol

now lets calculate the new molarities at total volume 100 + 100 = 200 ml = 0.200 L

[HPO4^2-] = 0.000918 mol / 0.2 L = 0.00459 M

[H2PO4^-] = 0.00066 mol / 0.2 L = 0.0033 M

So the conjugate acid base pair present is H2PO4^- and HPO4^2-

Now using the pka 2 of the phosphoric acid lets calculate the pH

pH= pka2 + log ([Base]/[acid])

pH = 7.59 + log [0.00459/0.0033]

pH= 7.73

Therefore pH= 7.73