Question

How many grams of NaH₂PO₄ (MW = 119.9) and how many mL of 1M NaOH would be needed to make 500 mL of a 0.2 M phosphate buffer at pH 7.5? The pka of phosphate is 7.21.

Answer 1

Using

pH = pKa + log ( moles of NaH₂PO₄/ moles of H₃PO₄)

Or, 7.5 = 7.21 + log(moles of NaH₂PO₄/moles of H₃PO₄)

Or, 0.29 = log( moles of NaH₂PO₄/ moles of H₃PO₄)

Or, ( moles of NaH₂PO₄ / moles of H₃PO₄) = 10^0.29 = 1.95

Let x mole NaH₂PO₄ and y mole NaOH is added.

So, (x/y) = 1.95

Total Molarity of phosphate buffer = 0.2 M

Hence, total moles = 500× 10^-3 × 0.2 = 0.1.

	NaH2PO4	NaOH	H3PO4
Before	X	y	O
Change	-y	-y	+y
After	x-y	0	y

So, ( x - y) +y = 0.1

Or, x = 0.1

Hence,

(x-y) /y = 1.95

Or, x - y = 1.95 y

Or, x = 2.95 y

Or, y = (x/2.95)

Or, y = (0.1/2.95) = 0.0338

Now from eq.1.

Therefore moles of NaH₂PO₄ added = 0.1

Mass of NaH₂PO₄ = mole × molar mass

= 0.1×119.9 = 11.99 g

and volume of 1 M NaOH needed = (moles of NaOH/ molarity)

= (0.0338/1)

= 0.0338 L

= 33.8 mL.

So, take 11.99 g NaH₂PO₄ add 33.8 mL 1 M NaOH , then add distilled water upto volume 500 mL.