# Question 50bb0

Jun 22, 2017

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.

Jun 22, 2017

$30$ $\text{g NaOH}$, if I get what you're asking:

#### Explanation:

I'll assume that you mean that the desired solution is 30% by mass $\text{NaOH}$.

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

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

and the remaining $70$ $\text{g}$ 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 $\text{^"o""C}$, and the $y$-axis represents the grams of $\text{NaOH}$ able to be dissolved in $100$ $\text{g}$ water.

Since there are $70$ $\text{g}$ water in the solution (found earlier), at ${0}^{\text{o""C}}$ (where according to the curve the solubility is roughly $\left(44 \textcolor{w h i t e}{l} \text{g NaOH")/(100color(white)(l)"g H"_2"O}\right)$) there would be only

70cancel("g H"_2"O")((44color(white)(l)"g NaOH")/(100color(white)(l)cancel("g H"_2"O"))) = 30.8 $\text{g NaOH}$

that dissolve, just barely above the $30$ $\text{g}$ value found earlier.

We can conclude that the solubility of sodium hydroxide in $70$ $\text{g}$ water is roughly $30.8$ $\text{g}$.

Therefore, as long as the solution temperature is above ${0}^{\text{o""C}}$, 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 ${0}^{\text{o""C}}$, we can't necessarily expect all the $\text{NaOH}$ 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 $\text{NaOH}$ then.