Question

what is pH of 15.0g KH2PO4 + 27.0 g Na2HPO4 dissolved in H2O plus enough H2O to fill to 1L

Answer 1

Let us calculate first concentrations(moles/L) of KH2PO4 and Na2HPO4,

i) Molecular weight of KH2PO4 is 136.1 g/mol

SO, conc. of KH2PO4 = mass of KH2PO4 / Mol. wt. of KH2PO4

= 15.0 / 136.1

conc. of KH2PO4 = 0.1102 moles/L

ii) Molecular weight of Na2HPO4 = 142 g/mol

So, conc. of Na2HPO4 = 27 / 142

= 0.1901 moles/L

Both KH2PO4 and Na2HPO4 disociates completely and hence concentrations of H2PO₄^- and HPO₄^2- .are same as that of their salts

hence, [H2PO₄^-] =0.1102 moles/L and [HPO₄^2-] = 0.1901 moles/L

Here we H2PO₄^- have as acid with the conjugate base HPO₄^2- .

H2PO₄^- <----------->H2PO₄^- + H⁺ .

Now pKa2 i.e. for second ionization constant for H2PO₄^- is 7.20

pKa₂ = 7.2

Now by Handerson equation,

pH = pKa2 + log{[Base or salt] / [Acid]}

pH = pKa2 + log{[HPO₄^2-] / [H2PO₄^-]}

pH = 7.20 + log(0.1901 / 0.1102)

pH = 7.20 + log(1.7250)

pH = 7.20 + 0.24

pH = 7.44 .

Note Na+ and K+ are spectator ions and do not contributes in pH of solution.