calc (1 Viewer)

[dchait]

Hi Guys,

How do i do this:

A 0.13mol L-1 NaOH solution was required for a titration

1. Calculate OH- concentration of solution
2. Calculate H+ concentration of solution
3. Calculate pH of solution

 

[Jumbo Cactuar]

Hi dchait,

NaOH completely dissociates in water.

So [OH-] = [NaOH]
[OH-] = 0.13 M

The equation for the ionic product of water is;

[H+][OH-] = K_w = 1.008*10^-14 @ 25C
[H+] = 0.77 fM (that's femtomolar!)

pH = - log [H+]
pH = -log 0.7753846...*10^-15
pH = 13.1

 

[Dreamerish*~]

Hi Guys,

How do i do this:

A 0.13mol L-1 NaOH solution was required for a titration

1. Calculate OH- concentration of solution
2. Calculate H+ concentration of solution
3. Calculate pH of solution

1. The reaction of NaOH in water is:

NaOH_(aq) → Na⁺_(aq) + OH^-_(aq)

NaOH is a strong base, which means the reaction goes to completion.

0.13 mol L^-1 solution was used, therefore 0.13 mol of NaOH dissociates to 0.13 mol of Na⁺ and also 0.13 mol of OH^-.

Hence the concentration of OH^- in solution is 0.13 mol L^-1

[OH^-1] = 0.13 mol L^-1

2. At standard room temperature, K_w = [H⁺] × [OH^-1] = 1 × 10^-14

[H⁺] × 0.13 = 1 × 10^-14

Hence [H<sup>+[/sub]] = 7.69 × 10-14 mol L-1

3. pH = -log₁₀[H+]

pH = -log₁₀(7.69 × 10-14)

pH ≈ 13</sup>