javeya

A particular household ammonia solution (d= 0.97g/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 = 11.55? ------------------------- i am gettig 5.3 x 10^-10 is that rite...

first i did was ph = 11.50 PH= 14-11.50 = 2.5 10^-2.5 = 0.0032 M Volume - 0.65 L d=0.97g/mL Mass %= 6.8% 1000mL= 1L d=m/v m=d X v m= 0.97g/mL X 1000mL m=970g mNH3 = 970g X 0.068=65.96g of NH3 MM of NH3 = 14g/mol + 3 X 1g/mol=17g/mol nNH3 = 65.96g/17g/mol...