Calculate the ph during the titration of 10.00 ml of 0.400 m hypochlorous acid with 0.500 m naoh. first what is the initial ph (before any naoh is added)? the ka for hocl is 3.0 x 10-8 m.

3.96

Explanation :

Set up an ICE table

       HClO + H₂O ⇌ H₃O⁺ + ClO⁻

I:     0.400                   0          0

C:       -x                      +x        +x

E: 0.400-x                    x          x

Kₐ = {[H₃O⁺][ClO⁻]}/[HClO] = 3.0 × 10⁻⁸        Substitute values

(x × x)/(0.400-x) = 3.0 × 10⁻⁸                        Combine like terms

x²/(0.400-x)                                                  Check that x ≪ 0.400

0.400/(3.0 × 10⁻⁸) = 1.3 × 10⁷ > 400             x ≪ 0.400; Ignore x

x²/0.400 = 3.0 × 10⁻⁸                                   Cross-multiply

x² = 0.400 × 3.0 × 10⁻⁸                                Complete the multiplication

x² = 1.20 × 10⁻⁸                                             Take the square root of both sides

x = [H₃O⁺] = 1.10 × 10⁻⁴                                 Take the negative log of each side

-log[H₃O⁺] = pH = -log(1.10 × 10⁻⁴)               Complete the operation

pH = 3.96

i do not get it   at all