Question

1. How would you make 100 mL of a carbonic acid buffer at 0.5 M and pH = 6.0 using 1.0 M NaHCO3 and either 1.0 M NaOH or 1.0 M HCl and water? (please Show your work.)

Answer 1

Answer:

We know that P^H = P^ka + log [salt] / [acid]

Here the Salt is NaHCO₃

The acid is HCl

We know that P^ka for the weak acid is 4.75

P^H = 4.75 + log (1) / (1)

P^H = 4.75

[H⁺] = 10^-PH

[H⁺] = 10^-4.75

[H⁺] = 10^0.25 x 10^-5

[H⁺] = 10^0.25 x 10^-5

[H⁺] = 1.778 x 10^-5 M

M₁V₁ = M₂V₂

M₁ =  1.778 x 10^-5      V₁ = ?

M₂ = 0.5 M V₂ = 100 mL

M₁V₁ = M₂V₂

1.778 x 10^-5 x V₁ = 0.5 x 100

V₁ = (0.5 x 100) / 1.778 x 10^-5

V₁ = 28.121 x 10⁵ mL

HCl diluted with above volume of Sodium bicarbonate solution.