Question

A buffer is made from 3.7 ?? 10?1 M of HC2H3O2 and 2.9 ?? 10?1 M of NaC2H3O2

a. What is the pH?

b. If you added 3.0 ?? 10?3 moles of NaOH to 0.100 L of buffer what is the final pH. (assume that any change in volume is negligible.)

Answer 1

Answer –

We are given, [HC₂H₃O₂] = 3.7*10^-1 M , [NaC₂H₃O₂]= [C₂H₃O₂^-] = 2.9*10^-1 M

a)pH of this buffer solution –

From the given data we need to use Henderson Hasselbalch equation-

We know formula –

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

so we need to first calculate the pKa for HC₂H₃O₂, Ka value is 1.75*10^-5

so, pKa = -log Ka

              = -log 1.75*10^-5

              = 4.756

Now plugging the value in the formula

pH = 4.756 + log 2.9*10^-1 M / 3.7*10^-1 M

     = 4.756+ (-0.1058)

    = 4.65

b) When we added 3.0*10^-3 moles of NaOH in the 0.100 L then pH is –

When we added NaOH means it gets reacted with acid and added in the base

So moles of acid, HC₂H₃O₂ = 3.7*10^-1 M * 0.100 L = 0.037

Moles of conjugate base , C₂H₃O₂^- = 2.9*10^-1 M * 0.100 L =0.029

New moles of acid and conjugate base after added NaOH

Moles of HC₂H₃O₂ = 0.037 moles - 3.0*10^-3 moles = 0.034

Moles of C₂H₃O₂^- = 0.029 moles + 3.0*10^-3 moles = 0.032

So new molarity is

[HC₂H₃O₂] = 0.034 moles / 0.100 L = 0.34 M

[C₂H₃O₂^-] = 0.032 /0.100 L = 0.32 M

Now again using the Henderson Hasselbalch equation-

pH = 4.756 + log 0.32 /0.34

      = 4.756 + (-0.0263)

      = 4.73