Question

what volume of 0.15 M sodium hydrogen oxalate must be added to 20.00 ml of 0.25 M sodium oxalate to produce a solution with a ph of 4.80?

Answer 1

Sodium hydrogen oxalate = 0.15M

let be the volume of Sodium hydrogen oxalate = V L

number of moles of Sodium hydrogen oxalate = 0.15M x V L = 0.15V mole

Sodium oxalate = 20.00 mL of 0.25M

number of moles of Sodium oxalate = 0.25M x 0.0200L = 0.005 moles

PH = 4.80

Ka2 of hydrogen oxalate = 6.4x10^-5

Ka= 6.4x10^-5

-log(Ka) = -log(6.4x10^-5)

Pka = 4.19

PH = Pka + log[salt/acid]

4.80 = 4.19 + log(0.005 / 0.15V)

0.61 = log( 0.005 /0.15V)

0.005/ 0.15V= 10^0.61

0.005/ 0 15V = 4.074

0.005 = 0.6111V

V= 0.0082

Volume of Sodium hydrogen oxalate = 0.0082L = 8.2 mL

volume of Sodium hydrogen oxalate = 8.2 mL