Question

PO, and of K2HPO, PM phosphate buffer for KH2PO4 and 1/4.10 YM 15. (LO 12) Calculate the mass of KH2PO4 and of ko required to prepare 100 mL of a 150 mm phospha at pH 7.2. The relevant acid dissociation is te H2PO4 = H+ + HPO2-nas and the pk, = 7.2. The formula weights are 136.09 g/mol for KH2PO4 and 174.18 g/mol for K2HPO4.

Answer 1

We have Henderson's equation for acidic buffer. pH = pKa + log [ Salt] / [ Acid]

We have, p H = 7.20 , pKa = 7.20 , [Salt] = [ K₂HPO₄ ] and [ Acid ] = [ KH₂PO₄ ]

Placing all these values in above equation, we get

7.20 = 7.20 + log [ K₂HPO₄ ] / [ KH₂PO₄ ]

$\therefore$ log [ K₂HPO₄ ] / [ KH₂PO₄ ] = 7.20 - 7.20 = 0.00

Taking antilog on both sides we get [ K₂HPO₄ ] / [ KH₂PO₄ ] = 1.

$\therefore$ [ K₂HPO₄ ] = [ KH₂PO₄ ] $\rightarrow$ (1)

We have given ,  [ K₂HPO₄ ] + [ KH₂PO₄ ] = 0.15 M $\rightarrow$ (2)

Put [ K₂HPO₄ ] = [ KH₂PO₄ ] in equation 2.

$\therefore$   [ K₂HPO₄ ] + [ K₂HPO₄ ] = 0.15 M

2  [ K₂HPO₄ ] = 0.15 M

[ K₂HPO₄ ] = 0.15 M / 2 = 0.075 M

$\therefore$  [ K₂HPO₄ ] = [ KH₂PO₄ ] = 0.075 M

We can write , No. of moles of K₂HPO₄ = No. of moles of KH₂PO₄

We have, relation , Molarity = No. of moles of solute / volume of solution in L

$\therefore$ No. of moles of K₂HPO₄ = 0.075 mol / L $\times$ 0.100 L = 0.0075 mol

$\therefore$ No. of moles of KH₂PO₄ = 0.075 mol / L $\times$ 0.100 L = 0.0075 mol

We have, No. of moles = Mass/ Molar mass

Mass of K₂HPO₄ = No. of moles of K₂HPO₄ ( Molar mass) = 0.0075 mol $\times$ 174.18 g / mol = 1.306 g

Mass of KH₂PO₄ = No. of moles of KH₂PO₄ ( Molar mass) = 0.0075 mol $\times$ 136.09 g / mol = 1.021 g

ANSWER : Mass of K₂HPO₄ = 1.306 g & Mass of KH₂PO₄ = 1.021 g