Question

What volume of 0.100 M HClO4 solution is needed to neutralize 60.00 mL of 8.80×10−2 M NaOH?
What volume of 0.134 M HCl is needed to neutralize 2.75 g of Mg(OH)2?
If 26.8 mL of AgNO3 is needed to precipitate all the Cl− ions in a 0.770-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution?
If 45.0 mL of 0.104 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)

the reaction is

HCl04 + NaOH ---> NaCl04 + H20

we can see that

moles of HCl04 required = moles of NaOH

also

moles = molarity x volume

so

0.1 x V = 8.8 x 10-2 x 60

V = 52.8

so

52.8 ml of HCl04 is needed

2)

we knoow that

moles = mass / molar mass

so

moles of Mg(OH)2 = 2.75 / 58 = 0.0474

now

the reaction is

Mg(OH)2 + 2HCl ---> MgCl2 + 2H20

we can see that

moles of HCl= 2 x moles of Mg(OH)2

so

moles of HCl = 2 x 0.0474 = 0.0948

now

volume = moles x 1000/ molarity

so

volume = 0.0948 x 1000 / 0.134

volume = 707.67

so

707.67 ml of HCl is needed

3)

moles of KCl = 0.77 x 10-3 / 74.55 = 1.033 x 10-5

so

moles of Cl- = 1.033 x 10-5

now

Ag+ + Cl- ---> AgCl

we can see that

moles of Ag+ = moles of Cl- = 1.033 x 10-5

so

moles of AgN03 = 1.033 x 10-5

noow

molarity = moles x 1000 / volume (ml)

so

molarity = 1.033 x 10-5 x 1000 / 26.8

molarity = 3.854 x 10-4

so

molarity of AgN03 is 3.854 x 10-4 M

4)

moles of HCl = 0.104 x 45 x 10-3

moles of HCl = 4.68 x 10-3

now

KOH + HCl = KCl + H20

so

moles of KOH = moles of HCl = 4.68 x 10-3

now

mass = moles x molar mass

so

mass of KOH = 4.68 x 10-3 x 56 = 0.2621

so

0.2621 grams of KOH is needed