Question

How many grams of `\text{NH}_4\text{OH}` do I need to make `400\ \text{mL}` of a `0.45\ \text{M}` solution of `\text{NH}_4\text{OH}`...?

Answer 1

`"6.3072 g"`

Explanation:

`"Molarity" = "Moles of solute"/"Volume of solution (in litres)"`

`"0.45 M" = "n"/"0.4 L"`

`"n = 0.45 M × 0.4 L = 0.18 mol"`

You need `"0.18 mol"` of `"NH"_4"OH"`

Molar mass of `"NH"_4"OH"` is `"35.04 g/mol"`

Mass of solute `= 0.18 cancel"mol" × "35.04 g"/cancel"mol" = "6.3072 g"`

Answer 2

To make the solution you take 10.0 ml of concentrated ammonia solution and make up to 400. ml with distilled water.

Explanation:

The answer by @Junaid Mirza is perfectly correct in theory but if you wanted to make such a solution from a practical point of view then you can't make it that way.

The problem is that ammonium hydroxide does not exist as a discrete substance so you can't weigh out a specific amount and make it up into a solution.

Ammonia gas `sf(NH_3)` is very soluble in water where the following equilibrium exists:

`sf(NH_(3(g))+H_2O_((l))rightleftharpoonsNH_(4(aq))^(+)+OH_((aq))^(-))`

The position of equilibrium lies well to the left so such a solution contains only a very small amount of `sf(NH_4^+)` and `sf(OH^(-))` ions.

It is better described as aqueous ammonia `sf(NH_(3(aq)))`.

The term "ammonium hydroxide" was used historically to describe such a solution but the term is largely obsolete, such that you may only find it in old text books or outdated syllabi.

Ammonia generally is sold as a concentrated solution which is about 18 M.

To make the solution you require you can use the dilution expression:

`sf(C_1V_1=C_2V_2)`

`:.` `sf(18xxV_1=0.45xx400.)`

`:.` `sf(V_1=(0.45xx400.)/18=10color(white)(x)"ml")`

So take out 10. ml of the concentrated ammonia with a pipette and make up to the 400. ml mark with distilled water.