Question

How many moles of sodium hydroxide would have to be added to 250 mL of a 0.357 M hypochlorous acid solution, in order to prepare a buffer with a pH of 7.640?

Answer 1

NaOH required for...

HClO solution to get pH =7.64

HClO + NaOH = Na+ + ClO- and H2O will be present

ClO- and HClO areconjugate base/weak acd}

A buffer is any type of substance that will resist pH change when H+ or OH- is added.

This is typically achieved with equilibrium equations. Both type of buffer will resist both type of additions.

When a weak acid and its conjugate base are added, they will form a buffer

The equations:

The Weak acid equilibrium:

HA(aq) <-> H+(aq) + A-(aq)

Weak acid = HA(aq)

Conjugate base = A-(aq)

Neutralization of H+ ions:

A-(aq) + H+(aq) <-> HA(aq); in this case, HA is formed, H+ is neutralized as well as A-, the conjugate

Neutralization of OH- ions:

HA(aq) + OH-(aq) <-> H2O(l) + A-(aq) ; in this case; A- is formed, OH- is neutralized as well as HA.

use henderson hasslebach

pH = pKa + log(ClO- / HClO)

pKa = 7.46

pH = pKa + log(ClO- / HClO)

7.640= 7.46+ log(ClO- / HClO)

10^(7.640 -7.46) = (ClO- / HClO)

(ClO- / HClO) = 1.5135

initially

mmol of HClO = MV = 0.357*250 = 89.25 mmol

mmol of ClO-= 0

after adding

mmol of NaOH = x

mmol of HClO = 89.25 - x

mmol of ClO-= 0 + x

and we know

(ClO- / HClO) = 1.5135

x /(89.25 -x) = 1.5135

x = (1.5135)(89.25 ) - 1.5135x

(1+1.5135)x = 135.07

x = 135.07 / 2.5135 = 53.73 mmol of NaOH required

mass = mmol*MW = 53.73*40 = 2149.2 mg = 2.15 g of NaOH