Question #50bb0

2 Answers
Jun 22, 2017

Answer:

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

Answer:

#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.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.