Question

If you began with 5.5 mL of 0.20 M NaHCO₃, how many mL of 0.10M HCl would you need to add to get a pH 7 buffer? Given that carbonic acid, H₂CO₃, has a pK_a1 = 6.37 and a pK_a2 = 10.33

Answer 1

given NaHC03 and HCl

the reaction is

NaHC03 + HCl ----> NaCl + H2C03

now

NaHC03 and H2C03 form a buffer

we know that

for a buffer

pH = pKa + log [ conjugate base / salt]

so

pH = pKa1 + log [NaHC03 / H2C03]

7 = 6.37 + log [ NaHC03 / H2C03]

[ NaHC03 / H2C03] = 4.2658

we know that

moles = molarity x volume

as the final volume is same for both, they cancel out

moles of NaHC03 / moles of H2C03 = 4.2658

now

we know that

moles = molarity x volume

so

initial moles of NaHC03 = 0.2 x 5.5 x 10-3 = 1.1 x 10-3

let y moles of HCl be added

so from the reaction we get

moles of NaHC03 reacted = moles of HCl added = y

moles of NaHC03 remained = 1.1 x 10-3 - y

moles of H2C03 formed = moles of NaHC03 reacted = y

now

moles of NaHC03 / moles of H2C03 = 4.2658

( 1.1 x 10-3 - y ) / y = 4.2658

y = 2.089 x 10-4

so

2.089 x 10-4 moles of HCl is required

now

volume (L)= moles / molarity

=2.089 x 10-4 / 0.1

= 2.089 x 10-3

so

volume = 2.089 x 10-3 L

volume = 2.089 ml

so

2.089 ml of HCl should be added