Question

1- If 27.0 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.795-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution?

2- If 45.7 mL of 0.110 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

Answer 1

1)

moles of KCl = 0.795 x 10-3 / 74.55 = 1.0664 x 10-5

so

moles of Cl- = 1.0664 x 10-5

now

Ag+ + Cl- ---> AgCl

we can see that

moles of Ag+ = moles of Cl- = 1.0664 x 10-5

so

moles of AgN03 = 1.0664 x 10-5

noow

molarity = moles x 1000 / volume (ml)

so

molarity = 1.0664 x 10-5 x 1000 / 27

molarity = 3.95 x 10-4

so

molarity of AgN03 is 3.95 x 10-4 M

2)

we know that

moles = molarity x volume (ml) / 1000

so

moles of HCl = 0.11 x 45.7 x 10-3

moles of HCl = 5.027 x 10-3

now

KOH + HCl = KCl + H20

so

moles of KOH = moles of HCl = 5.027 x 10-3

now

mass = moles x molar mass

so

mass of KOH = 5.027 x 10-3 x 56 = 0.2815

so

0.2815 grams of KOH is needed