Question

Exercise 16.34

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 625 mL of a solution with pH = 11.65?

Express your answer using two significant figures.

Answer 1

0.97 g/m2 mass + Solution given: The density NH3 The chemical reaction is OH + NH4 NH3 + H2O According to the chemical equation I mole of NH3 with produce of 0 My dao xide ion. Thea concentrations PH 11.05 pol = 2.95 ndle -

poh - log [OH-] 2.98 = - log [on 10-12.15) (64) 으 3 1:12 X163 M Cor) let the mass of solution a cogn Thus, the mass of NH3 (6.8%) = 6.89

Calculate the moles of ammonia mass (92 6.83 17.03 g/mole Molar Me SS 0.39 Mole of NH3 Now, Glew late the volume esing the equation Mass (oo 8 Volumes Density 0.97 g/m2 . . 103 m2 1000 m2 4 Ltr 108m2 103 = 0.1034 1000 Calculate the concentration solution Then, f 0.39 Mole M= mole Volume o.103 L M 3.78 M

Now, we can use the dilution Collalarte volume eara tion to Mive M2 V2 Vy M2 V2 My V2 (1.12 X 70 ) (0.6252) 3.78 MM Vio 0.000185 L 1000 m2 Ilon IL ro 0.000185 2= 1000 x 0,00018 = 0.185 m2