An industrial chemist studying bleaching and sterilizing prepares several hypoclorite buffers.

find the ph of 0.300 m hclo and 0.350 m naclo. (ka for hclo = 2.9 x 10-8)ph =

pH is 7.60

Explanation:

Let's think the equations:

HClO + H₂O  ⇄  ClO⁻  +  H₃O⁺

As every weak acid, we make an equilbrium

The salt is dissociated in solution

NaClO → Na⁺  +  ClO⁻

                  HClO    +    H₂O    ⇄    ClO⁻    +     H₃O⁺

Initially       0.3m                            0.35m

We have the moles of acid, and the moles of conjugate base.

Reacts          x                                    X                  X

Some amount has reacted, so I obtained (in equilibrium) the moles of base + that amount, and the same amount for H₃O⁺ (ratio is 1:1)

HClO    +    H₂O    ⇄    ClO⁻    +     H₃O⁺

0.3 - x                        0.35 + x           x

Let's make the expression for Ka

Ka = [ClO⁻] . [H₃O⁺]  /  [HClO]

(we don't add water, because it is included in Ka)

2.9x10⁻⁸ = (0.35+x).x  / (0.3-x)

Ka is in order of 10⁻⁸, I can assume that 0.3-x is 0.3  and 0.35 +x =0.3

2.9x10⁻⁸ = (0.35)x  / (0.3)

(2.9x10⁻⁸ .  0.3) /0.35 = x

2.48x10⁻⁸ = x

This is [H₃O⁺]

For pH = - log [H₃O⁺]

pH = 7.60