What is the pH of the buffer formed by mixing 50 mL of 0.2 M NaH2PO4 with 50 mL of 0.12 M NaOH. ( Ka1(H3PO4)= 7.11x10-3 Ka2(H3PO4)= 6.32x10-8 Ka3(H3PO4)= 4.5x10-13 )

Answer 1

An alternative method to arrive at the same conclusion as J.R.S:

What is the pH of the buffer formed by mixing 50 mL of 0.2 M NaH2PO4 with 50 mL of 0.12 M NaOH. ( Ka1(H3PO4)= 7.11x10-3 Ka2(H3PO4)= 6.32x10-8 Ka3(H3PO4)= 4.5x10-13 )

OH^- must be the base, so H2PO4^- is the acid:

H2PO4^-+OH^- ↔ H2O + HPO4^2-

So, the henderson equation is:

pH=pKa+log([HPO₄^2-]/[H₂PO₄^-])

pKa is that of the acid, so of H2PO4-, because this buffer contains:

weak acid: H2PO4-

conj. base: HPO42-

Now consider this equation below:

pH=pKa+log(mol HPO₄^2- + mol base/ mol H₂PO₄^- - mol base)

This equation takes the entire ICE table or whatever and puts it into a single calculation. This only works if moles NaOH is less than moles H2PO4- (the acidic buffer component), because if the buffer component is in excess, a buffer is formed. So, determine the moles of each of the 3 things:

initial moles HPO42- = 0

mol NaOH base added = (50mL/1) (0.12 mol / 1000mL) = 0.006 mol NaOH

mol H2PO4- initial = (50 mL NaH2PO4/ 1 ) (0.2 mol NaH2PO4 / 1000 mL NaH2PO4) (1 mol H2PO4- / 1 mol NaH2PO4)=0.01 mol H2PO4-

Substitute into the equation, using the pKa of H2PO4-, since that is the acidic component of the buffer:

pH=pKa+log(mol HPO42- + mol base/ mol H2PO4- - mol base)

pH=-log(6.32x10-8)+log(0+0.006 /0.01-0.006 )

pH=7.38

Answer 2

Reactions:

NaH₂PO₄ + NaOH ==> Na₂HPO₄ + H₂O

Na₂HPO₄ + NaOH ==> Na₃PO₄ + H₂O

Let's determine which of these reactions are taking place.

Initial moles NaH2PO4 = 50 ml x 1 L/1000 ml x 0.2 mol/L = 0.01 mols

moles NaOH added = 50 ml x 1 L/1000 ml x 0.12 mol/L = 0.006 mols

Using an ICE table, we have...

NaH₂PO₄ + NaOH ==> Na₂HPO₄ + H₂O

0.01...............0.006.............0................Initial

-0.006........-0.006............+0.006...........Change

0.004.............0..................0.006............Equilibrium

Only the first reaction above takes place, so the acid we are interested in is NaH₂PO₄ and the Ka of interest is Ka₂ = 6.32x10^-8

Using the Henderson Hasselbalch equation, pH = pKa + log [salt]/[acid]

pKa = -log Ka = -log 6.32x10^-8 = 7.20

pH = 7.20 + log [Na₂HPO₄] / [ NaH₂PO₄]

pH = 7.20 + log (0.06/0.04) = 7.20 + 0.18

pH = 7.38