The Foundations of Chemistry

Similar relationships can be written for a solution of a base. Because 1 eq of acid always

reacts with 1 eq of base, we may write

Number of equivalents of acidnumber of equivalents of base

so

LacidNacidLbaseNbase or mLacidNacidmLbaseNbase

EXAMPLE 11-12 Volume Required for Neutralization

What volume of 0.100 NHNO 3 solution is required to neutralize completely 50.0 mL of a

0.150 Nsolution of Ba(OH) 2?

Plan

We know three of the four variables in the relationship

mLacidNacidmLbaseNbase, and so we solve for mLacid.

Solution

_?_ m Lacid75.0 mL of HNO 3 solution

You should now work Exercise 48.

In Example 11-13 let us again solve Example 11-7, this time using normality rather

than molarity. The balanced equation for the reaction of H 2 SO 4 with Na 2 CO 3 , inter-

preted in terms of equivalent weights, is

H 2 SO 4 
Na 2 CO 3 88nNa 2 SO 4 
CO 2 
H 2 O

1 mol 1 mol 1 mol 1 mol 1 mol

2 eq 2 eq 2 eq 2 eq 2 eq

98.08 g 106.0 g

So, 1 eq Na 2 CO 3 53.0 g

EXAMPLE 11-13 Standardization of Acid Solution

Calculate the normality of a solution of H 2 SO 4 if 40.0 mL of the solution reacts completely

with 0.364 gram of Na 2 CO 3.

Plan

We refer to the balanced equation. We are given the mass of Na 2 CO 3 , so we convert grams

of Na 2 CO 3 to equivalents of Na 2 CO 3 , then to equivalents of H 2 SO 4 , which lets us calculate

the normality of the H 2 SO 4 solution.



g

p

N

re

a

s

2

e

C

n

O

t

^3 88n eq

p

N

re

a

s

2

e

C

nt

O 3

 88n 

eq

ne

H

ed

2 S

e

O

d

^4 88n eq H

L

(^2) SO 4
50.0 mL0.150 N

0.100 N
mLbaseNbase

Nacid
The conversion factors needed to
convert liters to milliliters on each side
of this equation will cancel.
414 CHAPTER 11: Reactions in Aqueous Solutions II: Calculations
By definition, there must be equal
numbers of equivalents of all reactants
and products in a balanced chemical
equation.