Question

Consider how best to prepare one liter of a buffer solution with pH = 10.93 using one of the weak acid/conjugate base systems shown here.

Weak Acid	Conjugate Base	Ka	pKa
HC2O4-	C2O42-	6.4 x 10-5	4.19
H2PO4-	HPO42-	6.2 x 10-8	7.21
HCO3-	CO32-	4.8 x 10-11	10.32

How many grams of the sodium salt of the weak acid must be combined with how many grams of the sodium salt of its conjugate base, to produce1.00 L of a buffer that is 1.00 M in the weak base?

grams sodium salt of weak acid =

grams sodium salt of conjugate base =

Answer 1

Sol . As pKa value of HCO₃^- / CO₃^2- system is 10.32 which is near to pH 10.93 . So , this weak acid / conjugate base system is preferred to prepare 1 L of buffer with pH 10.93  

Now , Molarity of weak acid = [HCO₃^-] = 1.00 M  

So, Using Henderson - Hasselbalch equation ,  

pH = pKa + log( [CO₃^2-] / [HCO₃^-] )  

10.93 = 10.32 + log ( [CO₃^2-] / 1.00 )  

[CO₃^2-] / 1.00 = 10^0.61 = 4.0738

So , [CO₃^2-] = 4.0738 M

Now , Molarity of sodium salt of HCO₃^- = [NaHCO₃] = 1.00 M

and , Molarity of sodium salt of CO₃^2- = [Na₂CO₃] = 4.0738 M

As Volume of buffer = 1 L

So , moles of NaHCO₃ = 1.00 × 1 = 1 mol  

moles of Na₂CO₃ = 4.0738 × 1 = 4.0738 mol

As Molar mass of NaHCO₃ = 84.007 g/mol

So , Mass of NaHCO₃ = Moles of NaHCO₃ × Molar mass of NaHCO₃ = 1 × 84.007 = 84.007 g  

Also , Molar mass of Na₂CO₃ = 105.989 g/mol  

So , Mass of Na₂CO₃ = Moles of Na₂CO₃ × Molar mass of Na₂CO₃ = 4.0738 × 105.989 = 431.778 g