Question

What is the amount of Na2HPO4 you need to add to a solution of 25mg NaH2PO4 in 100ml water
to obtain a pH=7.95?

Answer 1

m = 25 mg of NaH2PO4

m = ??? mg of Na2HO4

This will form a buffer:

Na2HPO4 ---> 2Na+ and HPO4-2

HPO4-2 <--> H+ and PO4-2

NaH2PO4 -->Na+ and H2PO4-

H2PO4- <--> H+ and HPO4-2

Therefore:

HPO4-2 <--> H+ and PO4-2

H2PO4- <--> H+ and HPO4-2

There is a common ion, a buffer!

Ka = 6.2 × 10⁻⁸

pKa = -log(KA)= -log( 6.2 × 10⁻⁸) = 7.207

This is best modelled with a buffer equatoin

pH = pKa2 + log(HPO3 /PO4)

If you want pH = 7.95 then

7.95 = 7.21 + log(conjugate-/acid)

0.74 = log(conjugate-/acid)

10^0.74 = conjugate-/acid

5.495 = HPO4- / H2PO4

[HPO4-2] = 5.495*[H2PO4-]

fin concnetration of H2PO4 (25 mg per 100 ml)

MW of NaH2PO4 = 119.98 g/gmol

find moles in 25 mg

mol = mass/MW = (25*10^-3) / (119.98 ) =2.0836*10^-4 moles of NaH2PO4

which will deliver 2.0836*10^-4 moles of H2PO4

M = mol/V

V = 100 ml or 0.1 L

M = (2.0836*10^-4 / (0.1L) = 0.002083

[HPO4-2] = 5.495*[H2PO4-]

[HPO4-2]  = 5.495*0.002083 = 0.011449

[HPO4-2]  = 0.011449 mol per liter

but we need mass, probably in mg so:

V = 0.1 L

[HPO4-2]  = 0.011449

mol = M*V = 0.011449*0.1 = 0.0011449 mol needed

mass = mol*MW

MW of Na2HPO4 = 141.96 g/gmol

mass = mol*MW = 0.0011449 * 141.96 = 0.1625 grams

mass in mg = 0.1625 g *1000 mg/g =162.5 mg are needed

m = 162.5 mg of Na2HPO4