What volumes of 0200 M HCOOH and 200 M NaOH would make 0800

What volumes of 0.200 M HCOOH and 2.00 M NaOH would make 0.800 L of a buffer with the same pH as a buffer made from 475 mL of 0.200 M benzoic acid and 25 mL of 2.00 M NaOH? Volume of HCOOH = L HCOOH Volume of NaOH = L NaOH

Solution

we know that

moles = molarity x volume (L)

so

moles of C6H5COOH = 0.2 x 0.475 = 0.095

moles of NaOH = 2 x 0.025 = 0.05

now

C6H5COOH + NaOH --> C6H5COONa + H20

moles of C6H5COOH reacted = moles of NaoH added = 0.05

moles of C6H5COONa formed = moles of NaOH added = 0.05

now

moles of C6H5COOH remaining = 0.095 - 0.05 = 0.045

now

pH = pKa + log [ C6H5COONa / C6H5COOH ]

so

pH = 4.202 + log [ 0.05 / 0.045]

pH = 4.2477

now

for HCOOH

pH = pKa + log [ HC00Na / HCOOH ]

4.24777 = 3.77 + log [ HCOONa / HCOOH ]

[HCOONa / HCOOH ] = 3

[HCOONa ] = 3 [HCOOH]

as the final volume is same for both

ratio of concentrations = ratio of moles

so

moles of HCOONa = 3 x moles of HCOOH

now

let the volume of NaOH be V

then volume of HCOOH = 0.45 - V

now

moles = conc x volume

so

moles of NaoH = 2 x V = 2V

moles of HCOOH = 0.2 x ( 0.45 - V) = 0.09 - 0.2V

now

HCOOH + NaOH ---> HCOONa + H20

moles of HCOOH reacted = moles of NaOH added = 2V

moles of HCOONa formed = moles of NaOH added = 2V

moles of HCOOH remaining = 0.09 - 0.2V - 2V

moles of HCOOH remainign = 0.09 - 2.2 V

now

moles of HCOONa = 3 x moles of HCOOH

so

2V = 3 x ( 0.09 - 2.2V)

V = 0.0314

so

volume of NaOH = 0.0314 L

volume of HCOOH = 0.45 - V = 0.45 - 0.0314 = 0.4186 L