Question

Calculate the pH of 1.00 L of a buffer that is 0.120 M HNO2 and 0.150 M NaNO2.

then, Calculate the pH of the buffer described above after the addition of 1.00 mL of 12.0 M HCl.

Answer 1

By using Henderson Hasselbach equation,

pH = pKa + log ([base] / [acid])

pKa of HNO2 = 3.4

pH = 3.4 + log (0.150 / 0.120)

= 3.4 + log (1.25)

= 3.4 + 0.096

pH = 3.49

If 1.0 mL of 12 M KCl is added:

No. of moles of H⁺ are added, n = M* V

= (12 mol/L) * (1 * 10^{-3 L)}

     = 12 * 10^-3 mol

This is the ammount of base that will be neutralized (converted to acid)

The total no. of moles of base in the original buffer = 0.15

The moles of acid = 0.12

By neutralizing 0.012 mols of the base and converting it to acid,

the new number of moles = 0.138 and 0.132

Now from H-H equation, the new pH is,

pH = pKa + log (base/acid)

pH = 3.40 + log (0.138 / 0.132 )

pH = 3.42