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

Answer: The molarity of base is 0.103 M

Step-by-step 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* 0.175* 29.50=1* M_2* 50.0\\\\M_2=(1* 0.175* 29.50)/(1* 50.0)=0.103M`

Hence, the molarity of base is 0.103 M

Answer 2

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