Question

You are instructed to create 300. mL of a 0.34 M phosphate buffer with a pH of 7.3. NaH₂PO₄ is the acid component of the buffer. Na₂HPO₄ is the base component of your buffer.

What is the molarity needed for the acid component of the buffer?

What is the molarity needed for the base component of the buffer?

How many moles of acid are needed for the buffer?

How many moles of base are needed for the buffer?

How many grams of acid are needed for the buffer?

How many grams of base are needed for the buffer?

Answer 1

The ratio of acid component : base component in the phosphate buffer with pH = 7.3 is calculated using

Handerson's- Hasselbalch's equation

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

substituting the given values we get,

log { [ Na H2 PO4 ] / [Na2HPO4} } = pH - pKa*...........*pKa of NaH2PO4 = 7.2

.........................................................= 7.3 - 7.2 = 0.1

& [ NaH2PO4] / [ Na2HPO4]..........= 1.2589

hence the molarity of acid component = 1.2589 M

................................base component = 1.000 M

But we are instructed to prepare 300ml. of o.34 M buffer solution-

This can be found out by determining the volumes of acid and base components needed separately using relation

M1V1 = M2V2

Thus volume of acid component .........= ( 300 x 0.34 ) / 1.2589

.............................................................= 81.0 ml

hence moles of acid component needed = (81.0 x 1.2589) / 1000

................................................................ = 0.102 mols

and weight of acid component ................ = (0.102 x 120)

................................................................. = 12.24 gms.

Similarly, mols of base component is calculated as = 0.102 mols

...................weight of base component .....................= 14.48 gms.