Question

Calculate the pH of a mixture containing 50 mL of 0.1 M NaH2POa and 150 mL of 0.1 M Na2HPO4. How many mL of 0.1 M H3POa should be added to the above buffer to lower the pH by one unit?

Answer 1

pH = pka + log [conjugate base] /[acid]

acid is NaH2PO4 whose pka is 7.2 . conjugate basse is Na2HPo4

conc of NaH2PO4 after adding = M x V = ( 0.1 x 50) / ( 50+150) = 0.025

conc of Na2HPo4 after mixing = ( 0.1 x 150) / ( 50+150) = 0.075

now pH = 7.2 + log ( 0.075/0.025)

= 7.68

if pH to be lowered by 1 unit

6.68 = 7.2 + log [Na2HPO4]/[NaH2PO4]

[Na2HPO4] = 0.3 [NaH2PO2]

Na2HPO4 moles = 0.3 x NaH2PO4 moles

Initially we had NaH2Po4 moles = M x V = 0.1 x 50/1000 = 0.005

Na2HPo4 moles = 0.1 x 150/1000 = 0.015

let H+ ( i.e H3PO4 moles added = m)

then NaH2PO4 moles = 0.005 + m

Na2HPO4 moles = 0.015-m

now ( 0.015-m) = 0.3( 0.005+m)

0.015-m = 0.0015+0.3m

m = 0.0104 = Mx V of H3PO4

0.0104 = 0.1 x V

Vol of H3PO4 = 0.104 L = 104 ml