Nvm. I got it.
It wasn't so hard after all :D
How many milliliters of a 15.0%, by mass solution of KOH(aq) ( d= 1.14g/mL ) are required to produce 27.0 L of a solution with pH = 11.70?
I'm really not sure how to do this, it's from Acids & Bases chapter.
Any help is appreciated, thanks. (:
2 answers
You want pH = 11.70; therefore, pOH = 14.0-11.70 = ??
Then find OH^- from pOH and (OH^-) = ?? M.
How many moles must you have if you want that molarity (OH%-). M x L = moles OH^-.
What is the molarity of the 15.0% KOH solution?
The mass of 1 L of solution is
1000 mL x 1.14 g/mL = yy grams.
How much of that is KOH?
yy x 0.15 = zz grams KOH.
How many moles KOH is that?
zz grams/molar mass KOH = mols KOH in 1 L which is that molarity.
Then MKOH soln x LKOH soln = moles
M KOH you have. L KOH you want. and you know how many moles you want in the 27.0 L. Find L KOH and convert to mL.
Then find OH^- from pOH and (OH^-) = ?? M.
How many moles must you have if you want that molarity (OH%-). M x L = moles OH^-.
What is the molarity of the 15.0% KOH solution?
The mass of 1 L of solution is
1000 mL x 1.14 g/mL = yy grams.
How much of that is KOH?
yy x 0.15 = zz grams KOH.
How many moles KOH is that?
zz grams/molar mass KOH = mols KOH in 1 L which is that molarity.
Then MKOH soln x LKOH soln = moles
M KOH you have. L KOH you want. and you know how many moles you want in the 27.0 L. Find L KOH and convert to mL.
1 answer