### acid-base-calculations

#### Calculations with pH

The most imporatnt reason to use p-values is that we often apply concentrations of very diluted solutions.

It is much easier to say:

**pH = 6 than [H**_{3}O^{+}] = 10^{-6} mol/l.

Please don't ever forget tjat a high p-value always automatically means a very small corresponding real value.

pOH = 9 (a rather high value) means a low concentration of OH^{-}-ions:

**pOH = 9 → [OH**^{-}] = 10^{-9} mol/l

Saying of writing something about concentrations, the unit mol/l (mol per liter) may not fail; using the p-value, there is no need for any unit.

But mind you: the p-value is based upon the well known unit mol/l

To remember: in water with a normal temperature (say 20 - 25ºC): pH + pOH = pK_{W} = 14

So, as soon as you know pH, you also know pOH.

An aqueous solution is called NEUTRAL when pH = pOH, at whatever temperature. That is the most important criterium for a neutral solution.
It is similar to say: the concentrations of H_{3}O^{+} and OH^{-} are equal.

Adding acid to a solution means that the pH-value becomes lower and the pOH-value increases.

Adding a base means a higher pH and a lower pOH.

The values of pH can also be broken number, like 3.4 and 10.7 e.d.
This can difficult the mathematical calculations.

For example: if the pH=3.5, then the concentration [H_{3}O^{+}] equals 10^{-3,5}mol/l.

But often we do not accept broken exponents.

You have to see and understand immediately in this example that the concentration must be between the values 10^{-3} and 10^{-4}mol/l (because the pH is between 3 en 4).

A calculator of course immediately gives you the solution, but even without the machine, you must be able to execute the calculation:

pH = 3.5 = 4 – 0.5 → -log[H_{3}O^{+}] = 4 - 0.5 → [H_{3}O^{+}] = 3* x *10^{-4} mol/l.

#### Calculations with K_{A} and K_{B}and with pK_{A} with pK_{B}

Exercise:
Calculate the pH of the following solutions:
- 0,1M HAc
- 0,1M NH
_{3}
- 0,1M HCl

Answer a):

0.1M HAc means: 0.1 mol acetic acid (CH_{3}COOH) was dissolved in water with an final volume of 1 liter.

A part of the acid molecules dissociate in H^{+} and Ac^{-}.

The amount of H^{+} (in water H_{3}O^{+}) determines the value of the pH (=-log[H_{3}O^{+}]).

We must know this amount of H^{+}, just like the strength/weakness of the acid, or: the K_{A}

We consult the table to see that K_{A }= 10^{-4} or pK_{A} = 4

We know that HAc is a weak acid. The value of x will be small compared to the [HAc]

or: x can be neglected.

So: 0.03 mol HAc dissociated into ions → [H_{3}O^{+}] = 0.03 = 3* x *10^{-2} mol/l → pH = 2-log3 = 1.5

Never forget:

**
The weaker an acid, the smaller K**_{A}

The stronger an acid, the bigger K_{A}

you can check this statement in table I.

Att.:

Strong acids and bases in the table do not have a K-value.