Question

(a) What volume of 0.115 M HClO4 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH? (b) What volume of 0.128 M HCl is needed to neutralize 2.87 g of Mg1OH22? (c) If 25.8 mL of an AgNO3 solution is needed to precipitate all the Cl- ions in a 785-mg sample of KCl (forming AgCl), what is the molarity of the AgNO3 solution? (d) If 45.3 mL of a 0.108 M HCl solution is needed to neutralize a solution of KOH, how many grams of KOH must be present in the solution?

Answer 1

Answer:

a) HClO4 Volume= 38.04mL

b) HCl Volume= 0.19 litters = 190mL

c) AgNO3 Molarity= 0.41 mol/l = 0.41 M

d) KOHmass = 0.275g

Explanation:

In both, the neutralization reactions (a, b and d) and in the single exchange reactions (d) the relationship between the moles of the reagents indicates how much is needed to complete the reaction.

The number of moles of a substance can be calculated as:

moles= Molarity* volume (liters)      >For solutions

or

moles= mass(g)/MM (g/mol)      > For pure substance

MM= Molecular Mass

a) moles of NaOH = moles of HClO4

NaOH Molarity*NaOH Volume= HClO4 Molarity* HClO4 Volume

HClO4 Volume= (NaOH Molarity* NaOH Volume)/HClO4 Molarity

HClO4 Volume= (0.0875M*50.00mL)/0.115M

HClO4 Volume= 38.04mL

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

Mg(OH)2  mass/ Mg(OH)2 MM = 2*HCl Molarity* HCl Volume

HCl Volume= (Mg(OH)2  mass/ Mg(OH)2 MM )/ 2*HCl Molarity

HCl Volume= (2.87g/58.32g/mol)/ 2*0.128mol/l

HCl Volume= 0.19 l =190ml

c) moles of KCl = moles of AgNO3

KCl mass/ KCl MM = AgNO3 Molarity* AgNO3 Volume

AgNO3 Molarity= (KCl mass/ KCl MM )/AgNO3 Volume

AgNO3 Molarity= (0.785g/74.55 g/mol))/0.0258 l

AgNO3 Molarity= 0.41 mol/l = 0.41 M

d) moles of KOH= moles of HCl

KOHmass/ KOH MM = HCl Molarity* HCl Volume

KOHmass = HCl Molarity* HCl Volume*KOH MM  

KOHmass = 0.108 (mol/l) *0.0453 l*56.11( g/mol)

KOHmass = 0.275g