Question

How many milliliters of 1.18 M NaOH must be added to 125 mL of 0.10 M NaH2PO4 to make a buffer solution with a pH of 6.90?

Answer 1

Consider a reaction of NaOH with NaH₂PO₄

NaH₂PO₄ + NaOH Na₂HPO₄

No of moles of NaH₂PO₄ = Molarity x volume of solution in L

= 0.1 mol / L x 0.125 L

= 0.0125 mol

Lets use ICE table

Concentration ( moles)	NaH2PO4	NaOH	Na2HPO4
Initial	0.0125	X	0
Change	-X	-X	+X
Final	0.0125 - X	0.0	X

pH of buffer solution is calculated by using Henderson's equation as

pH = pKa + log [Na₂HPO₄ ] / [NaH₂PO₄]

Here pKa= 7.198

Therefore,

6.90 = 7.198 + log [Na₂HPO₄ ] / [NaH₂PO₄]

log [Na₂HPO₄ ] / [NaH₂PO₄] = 6.90 - 7.198 = - 0.298

[Na₂HPO₄ ] / [NaH₂PO₄] = 10 ^-0.298 = 0.504

[Na₂HPO₄ ] = 0.504 x [NaH₂PO₄]

X = 0.504 ( 0.0125 - X )

= 0.0063 - 0.504 X

X + 0.504 X = 0.0063

1.504 X = 0.0063

X = 0.0063 / 1.504 = 0.00419 = No of moles of NaOH

We have, Molarity = No of moles / Volume of solution in L

Volume of 1.18 M  solution in L = No of moles / Molarity

= 0.00419 mol / 1.18 mol / L

= 0.00355 L

= 3.55 ml

ANSWER: Add 3.55 ml of NaOH to the 125 ml of 0.1 M NaH₂PO₄