First, what is the molarity of the household ammonia solution?
Density = 0.97 g/mL so a liter has a mass of
0.97 g/mL x 1000 mL = 970 grams.
How much of that is NH3? It is
970 g x 0.068 = xx grams NH3.
How many moles is that?
xx grams NH3/molar mass NH3 = yy moles NH3/L which makes it yy M.
What is the concn of NH3 in the new solution to be prepared? pH = 11.50 means pOH 2.5 and OH^- = ?? M. Use that and Kb for NH3 to calculate (NH3) for the 650 mL. (NH3) = zz M
Then mL old NH3 x yy M = 650 mL new NH3 x zz M. Solve for mL old NH3.
Check my thinking.
A particular household ammonia solution (d = 0.97 g/mL) is 6.8% NH3 by mass.
How many milliliters of this solution should be diluted with water to produce 650 mL of a solution with pH = 11.50?
1 answer
6 answers