Question #50bb0

2 Answers
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 "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:

http://chemicals.etacude.comhttp://chemicals.etacude.com

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.