posted by Calvin .
A particular household ammonia solution (d=0.97 g/mL) is 6.8% NH3 by mass. How many mililiters of this solution should be diluted with water to produce 625mL of a solution with pH = 11.65.
Is that 6.8% by mass (w/w) or by volume. I will assume it is w/w which means 6.8g/100 g soln. Convert 6.8g NH3 to moles (moles = grams/molar mass) and convert 100 g soln to volume (mass = volume x density). Then calculate M NH3 = moles/L soln. That will give you the molarity of the solution you have.
Next calculate the molarity of the soln of the diluted material.
You want pH = 11.65. Calculate pOH (pH + pOH = pKw = 14) then use
NH3 + H2O ==> NH4^+ + OH^-
Set up an ICE chart for NH3. Substitute OH^- needed and solve for NH3 initially. That will allow you to use mL 6.8%soln x M6.8%soln = 625 x M NH3 needed. Solve for mL6.8%soln.