Question

If 29.50 ml of 0.175m nitric acid neutralizes 50.0 ml of ammonium hydroxide, what is the molarity of the base?

Answer 1

The molarity of the  base ( ammonium hydroxide)   is calculated  as follows

find  the moles  of  Nitric  acid  =molarity  x  volume

=29.5  x  0.175 =5.163  moles  of nitric  acid

write the   equation  for  reaction

HNO3+ NH4OH  =NH4NO3  +H2O

by  use  of  mole  ratio  between  HNO3 to NH4NO3   which  is 1:1  the  moles  of  NH4OH is  also  5.163 moles

molarity =moles/volume

=  5.163 /50=0.103 M  is  the  molarity  of the  base

Answer 2

Answer: The molarity of base is 0.103 M

Explanation:

To calculate the concentration of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HNO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $NH_4OH$

We are given:

$n_1=1\\M_1=0.175M\\V_1=29.50mL\\n_2=1\\M_2=?M\\V_2=50.0mL$

Putting values in above equation, we get:

$1\times 0.175\times 29.50=1\times M_2\times 50.0\\\\M_2=\frac{1\times 0.175\times 29.50}{1\times 50.0}=0.103M$

Hence, the molarity of base is 0.103 M