Question

What is the pH at the half equivalence point when 50.0 mL of 0.200 M NaOH is titrated with 0.100 M HCl? The starting pH is 13.3.

Answer 1

`"pH" = 12.70`

Explanation:

The first thing that you need to do here is to calculate the volume of the hydrochloric acid solution needed to reach the half equivalence point of the titration.

Now, in order to reach the half equivalence point, you need to add enough strong acid to neutralize exactly half of the number of moles of strong base that you've started with.

`"NaOH"_ ((aq)) + "HCl"_ ((aq)) -> "NaCl"_ ((aq)) + "H"_ 2"O"_ ((l))`

The initial solution of sodium hydroxide contained

`50.0 color(red)(cancel(color(black)("mL solution"))) * "0.200 moles NaOH"/(10^3color(red)(cancel(color(black)("mL solution")))) = "0.0100 moles NaOH"`

This means that at the half equivalence point, the solution must contain

`1/2 * "0.0100 moles NaOH" = "0.00500 moles NaOH"`

In other words, you need to add `0.0500` moles of hydrochloric acid--sodium hydroxide and hydrochloric acid react in a `1:1` mole ratio--in order to reduce the number of moles of sodium hydroxide to half of what you started with.

Use the molarity of the hydrochloric acid solution to determine the volume of acid needed.

`0.00500 color(red)(cancel(color(black)("moles HCl"))) * (10^3 quad "mL solution")/(0.100color(red)(cancel(color(black)("moles HCl")))) = "50.0 mL solution"`

This means that the total volume of the solution at the half equivalence point will be

`"50.0 mL + 50.0 mL = 100.0 mL"`

Consequently, you can say that the concentration of hydroxide anions at the half equivalence point is equal to

`["OH"^(-)] = "0.00500 moles OH"^(-)/(100.0 * 10^(-3) quad "L") = "0.0500 M"`

As you know, the `"pH"` of an aqueous solution at `25^@"C"` is equal to

`color(blue)(ul(color(black)("pH" = 14 - "poH")))`

Since

`color(blue)(ul(color(black)("pOH" = - log(["OH"^(-)]))))`

you can say that you will have

`"pH" = 14 - (-log(["OH"^(-)]))`

`"pH" = 14 + log(["OH"^(-)])`

Plug in your value to find

`"pH" = 14 + log(0.0500) = color(darkgreen)(ul(color(black)(12.70)))`

I'll leave the answer rounded to two decimal places, but keep in mind that you have three sig figs for your values, so you should report the answer as

`"pH" = 12.699`