Phosphoric acid is a triprotic acid with the following pKa values: Phosphoric ac
ID: 497356 • Letter: P
Question
Phosphoric acid is a triprotic acid with the following pKa values:
Phosphoric acid is a triprotic acid with the following pK_a values: pK_a1 = 2.148 pK_a2 = 7.198 pK_a3 = 12.375 You wish to prepare 1.000 L of a 0.0200 M phosphate buffer at pH 7.540. To do this, you choose to mix the two salt forms involved in the second ionization, NaH_2PO_4 and Na_2PO_4, in a 1.000 L volumetric flask and add water to the mark. What mass of each salt will you add to the mixture? mass NaH_2PO_4 = mass Na_2HPO_4 = What other combination of phosphoric acid and/or its salts could be mixed to prepare this buffer? (Check all that apply).Explanation / Answer
for pH = 7.54,
apply buffer equation:
pH = pKa + log(conjugate base / weak acid)
since the pKa nearest to that pH is pKa2, use the second ionization of H3PO4
so:
pH = pKa2 + log(HPO4-2 / H2PO4-)
so...
Na2HPO4, NaH2PO4
substitute data
7.198 = 7.54 + log(Na2HPO4/NaH2PO4)
if V = 1 L, then
total mol = 0.02 *1 = 0.02 moles of Na2HPO4+NaH2PO4
ratio from buffer:
7.198 = 7.54 + log(Na2HPO4/NaH2PO4)
10^(7.198 - 7.54) = Na2HPO4/NaH2PO4
Na2HPO4/NaH2PO4 = 0.4549
Na2HPO4+NaH2PO4 = 0.02
so:
Na2HPO4 = 0.4549*NaH2PO4
0.4549*NaH2PO4 +NaH2PO4 = 0.02
NaH2PO4 = 0.02/(1+0.4549) = 0.013746 mol
so
Na2HPO4 = 0.4549*NaH2PO4
Na2HPO4 = 0.4549*0.013746 = 0.00625 moles
mass of Na2HPO4 = mol*MW = 0.00625 *141.96 = 0.88725 g of Na2HPO4
mass of NaH2PO4 = mol*MW = 0.013746 *119.98 = 1.6492 g of NaH2PO4
Q2.
we would require
HPO4-2 and H2PO4- in equilibirum so:
a) can't be since this is the first ionization
b) if proper ratio is satated, the pH could be achieved
c) if proper ratio is satated, the pH could be achieved
d)if proper ratio is satated, the pH could be achieved
e) can't be, this is third ionization