Question

Question #50bb0

Answer 1

30 grams of NaOH.

Explanation:

Generally, the definition of % concentration for acids and bases is grams per 100 grams.

In this example, a 30% solution can be written as 30% wt./wt., which means that there are 30 g of NaOH in 100 g of final solution.

So you need to weigh out 30 g of NaOH and dissolve it in enough water to get a total mass of 100 g.

Answer 2

`30` `"g NaOH"`, if I get what you're asking:

Explanation:

I'll assume that you mean that the desired solution is `30%` by mass `"NaOH"`.

If the solution (which I take to be only water and aqueous sodium hydroxide) were to have a mass of `100` `"g"`, a `30%` by mass solution would have

`(100color(white)(l)"g soln")(30% "NaOH") = color(red)(30` `color(red)("g NaOH"`

and the remaining `70` `"g"` would be water.

Further information:

The solubility of sodium hydroxide in water is seen by its solubility curve:

where the `x`-axis represents the temperature of the solution, in `""^"o""C"`, and the `y`-axis represents the grams of `"NaOH"` able to be dissolved in `100` `"g"` water.

Since there are `70` `"g"` water in the solution (found earlier), at `0^"o""C"` (where according to the curve the solubility is roughly `(44color(white)(l)"g NaOH")/(100color(white)(l)"g H"_2"O")`) there would be only

`70cancel("g H"_2"O")((44color(white)(l)"g NaOH")/(100color(white)(l)cancel("g H"_2"O"))) = 30.8` `"g NaOH"`

that dissolve, just barely above the `30` `"g"` value found earlier.

We can conclude that the solubility of sodium hydroxide in `70` `"g"` water is roughly `30.8` `"g"`.

Therefore, as long as the solution temperature is above `0^"o""C"`, our `30%` by mass solution should completely dissolve in the water, and thus there won't be any undissolved solute left over.

However, if the solution temperature is lower than `0^"o""C"`, we can't necessarily expect all the `"NaOH"` to dissolve. There would simply be too much solute to be dissolved in the water at those temperatures, as allowed by its solubility, and so there will be some undissolved `"NaOH"` then.