Answer on Question #50438, Chemistry, Other
Task:
Part I
a) Why are hydrogen ions NEVER found in an aqueous solution?
HCN(aq) + SO4-2(aq) → HSO4-(aq) + CN-(aq)
What is the Bronsted - Lowry acid in this equation?
What is the Bronsted - Lowry base in this equation?
What is the conjugate acid in this equation?
What is the conjugate base in this equation?
b) 2NH3 + Ag+ → Ag(NH3)2+
What is the Lewis acid in this equation?
What is the Lewis base in this equation?
Part II
Given H2SO4 is sulfuric acid, HNO3 is nitric acid, and H3PO4 is phosphoric acid, name the
following: HCl, H2SO3, HNO2, H3PO2, HNO4, H2SO5, HI.
Part III
Work these titration calculations:
a) A titration of 15.0 cm3 of household ammonia, NH3, required 38.70 cm3 of 8.0 M HCl.
Calculate the molarity of the ammonia.
b) What volume of 5.0 M HNO3 is required to neutralize 25.00 cm3 of a 2.0 M NaOH solution.
c) Calculate the volume of 0.55 M HNO3 necessary to neutralize 55.00 cm3 of 0.45 M KOH.
Answer:
Part I
a) Hydrogen ions never found in an aqueous solution because the attach to the water
molecules to form H3O+.
Bronsted - Lowry acid in the equation is HCN(aq), because it donates the proton.
Bronsted - Lowry base in this equation is SO4-2(aq), because it receives the proton.
Conjugate acid in this equation is SO4-2(aq).
Conjugate base in this equation is HCN(aq).
b) Lewis acid in this equation is Ag+.
Lewis base in this equation is NH3.
Part II
HCl – hydrochloric acid;
H2SO3 – sulforous acid;
HNO2 – nitrous acid;
H3PO2 – hypophosphorous acid;
Part III
a)
HNO4 – peroxynitric acid;
H2SO5 – peroxymonosulphuric acid;
HI – hydroiodic acid.
NH 4OH  HCl  NH 4Cl  H 2O

  CM  V
V
 ( NH 4OH )   ( HCl )
CM 
CM ( NH 4OH ) V ( NH 4OH )  CM ( HCl ) V ( HCl )
CM ( NH 4OH ) 
CM ( HCl ) V ( HCl )
V ( NH 4OH )
CM ( NH 4OH ) 
8  0.0387
 20.64 M
0.015
HNO3  NaOH  NaNO3  H 2O

  CM  V
V
 ( HNO3 )   ( NaOH )
CM 
b)
CM ( HNO3 ) V ( HNO3 )  CM ( NaOH ) V ( NaOH )
V ( HNO3 ) 
CM ( NaOH ) V ( NaOH )
CM ( HNO3 )
V ( HNO3 ) 
2  0.025
 0.01l  10 cm3
5
HNO3  KOH  KNO3  H 2O
 ( HNO3 )   ( KOH )
c)
CM ( HNO3 ) V ( HNO3 )  CM ( KOH ) V ( KOH )
V ( HNO3 ) 
CM ( KOH ) V ( KOH )
CM ( HNO3 )
V ( HNO3 ) 
0.45  0.055
 0.045 l  45 cm3
0.55
https://www.AssignmentExpert.com