A buffer solution is composed of1.4g KH2PO4 (136.09g/mol) and 5.7g Na2HPO4 (141.98g/mol)?

the ka =6.2*10^-8
a)what is the pka of the weak acid H2PO4 ion
b)what is ph of buffer solution

Answer:
pKa = -log Ka

Since Ka is given to you, plug and chug. p always means - log of (e.g. pH = -log(H)

pKa = - log (6.2 x 10^-8)

pKa = 7.21

In order to solve for pH, use the Henderson-Hasselbalch equation, or as my students like to remember it, the David Hasslehoff equation

pH = pKa + log([base]/[acid])


You need, technically, the volume that these species are dissovled in. Putting moles in will get you the correct answer, but was there a volume given, because the equation technically wants the concentrations of base and acid

let's assume 1 L (in reality, it won't matter because you are taking the ratio of the two, so if they volume isn't 1 L, just divde moles by the volume, then plug in. But the ratio of base/acid will be the same because they are in the same solution!)

1.4 grams KH2PO4 * 1mol/136.09 grams = 0.0103 moles/1 L = 0.0103 M

5.7 grams Na2HPO4 * 1mol/141.98 grams = 0.0401 moles/1 L = 0.0401 M

pH = pKa + log([base]/[acid])

the base is the species that is always 1H less than the other species. Since KH2PO4 has more H's than Na2HPO4 (this is 1 H less) then Na2HPO4 is the base

pH = 7.21 + log ([0.0401]/[0.0103])

pH = 7.21 + log(3.89)

pH = 7.21 + 0.59

pH = 7.80

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