Question

1. You make a solution of 0.464 moles HClO2 and 0.464 moles ClO2- dissolved in 1.00 L. What is the minimum number of moles of KOH would need to be added in order to 'overwhelm' this buffer?

2. You make a solution of 0.316 moles HClO2 and 0.464 moles ClO2- dissolved in 1.00 L. What is the minimum number of moles of KOH would need to be added in order to 'overwhelm' this buffer?

Answer 1

1. Let us first calculate pH of HClO₂/ClO₂^- (Hypochlorous acid-hypochlorate buffer)

As, 1L solution is prepared,

[HClO2] = 0.464 mol/L and [ClO2^-] = 0.464 mol/L

For Hypochlorous acid, Ka = 1.1 x 10^-2 .

Hence pKa = -log(Ka) = -log(1.1 x 10^-2) = 1.96

pKa = 1.96

By Handerson equation for acidic buffer,

pH = pKa + log{[salt]/[acid]}

pH = pKa + log{ [ClO₂^-] / [ HClO₂] }

pH = 1.96 + log{ [0.464] / [ 0.464] }

pH =1.96 + log(1)

pH = 1.96

pH of buffer formed is 1.96

Let us calculate [H⁺]

pH = -log[H⁺]………..(logarithmic form)

[H⁺] = 10^-pH…………..(Exponntional form)

[H⁺] = 10^-1.96

[H⁺] = 1.1 x 10^-2 M............. (Equimolar solution of acid and base used hence [H⁺] concentration is numerically equal to dissociation constant of acid)

At equivalence point equal concentration of [OH^-] will be present.

Hence [OH^-] =1.1 x 10^-2 M

It means to overwhelm buffer pH the minimum moles of KOH must be slightly more than 1.1 x 10^-2 M.

2) Now ,

[HClO2] = 0.316 mol/L and [ClO2^-] = 0.464 mol/L

With these new values and all other known values using Handerson eq.(1) let us calculate pH of this buffer

We have,

pH = pKa + log{[salt]/[acid]}

pH = pKa + log{ [ClO₂^-] / [ HClO₂] }

pH = 1.96 + log{ [0.316] / [ 0.464] }

pH =1.96 + log(0.68100

pH = 1.96 + (-0.1668)

pH = 1.79

[H⁺] = 10^-1.79

pH of this buffer is 1.79

Let us calculate [H⁺] =?

[H⁺] = 1.6 x 10^-2 M.

Hence at equivalence point,

[OH^-] =1.1 x 10^-2 M

To overwhelm buffer pH the minimum moles of KOH must be slightly more than 1.1 x 10^-2 M.