What volume of 0.101 M HNO3 is required to neutralize each of the following solutions?
(a) 12.5 mL of 0.525 M NaOH
mL
(b) 20.5 mL of 0.00455 M Ba(OH)2
mL
(c) 45.1 mL of 0.151 M NH3
mL
(d) 1.41 L of 0.107 M KOH
L
2007-05-07 21:56:32 · 3 answers · asked by kevin 1 in Science & Mathematics ➔ Chemistry
are you just going to keep on asking every homework problem you have?
2007-05-07 22:02:49 · answer #1 · answered by Sam 5 · 0⤊ 0⤋
Utilise always the following formula
n1V1 = n2V2
(where n are the normalities of the reactants anD V the volume)
a) Na OH and HNO3 are monovalent (normality 1)
12.5*0.525= 0.101* V (NO3H)
V (NO3H) = 12.5*0.525/0.101=65mL
b) here Ba (OH)2 is divalent so normality =0.00455*2= 0.0091
and 20.5*0.0091= 0.101V (NO3H)
V (NO3H) = 20.5*0.0091/ 0.101 =2.25mLNO3H
c) NH3 is monvalent
so 45.1*0.151=V (NO3H) *0.101=67.3mL NO3H
d) KOH is monovalent
so 1.41*0.107=0.101V (NO3H)
V(NO3H )= 1.41*0.107/0.101=1.494L
2007-05-07 22:16:19 · answer #2 · answered by maussy 7 · 0⤊ 0⤋
you could have asked my bro. he's genius in phy. , chem. and maths...
2007-05-07 22:02:29 · answer #3 · answered by namrata 2 · 0⤊ 0⤋