Question

It took 38.33 mL of 0.0944 M HCl to titrate (react completely with) the ammonia. What is the concentration of the original ammonia solution?

Answer 1

The question is incomplete, here is the complete question:

A 50.60 mL sample of an ammonia solution is analyzed by titration with HCl. The reaction is given below.

`NH_3(aq.)+H^+(aq.)\rightarrow NH_4^+(aq.)`

It took 38.33 mL of 0.0944 M HCl to titrate (react completely with) the ammonia. What is the concentration of the original ammonia solution?

Answer: The concentration of original ammonia solution is 0.0715 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 HCl

`n_2,M_2\text{ and }V_2` are the n-factor, molarity and volume of base which is
`NH_3`

We are given:

`n_1=1\\M_1=0.0944M\\V_1=38.33mL\\n_2=1\\M_2=?M\\V_2=50.60mL`

Putting values in above equation, we get:

`1* 0.0944* 38.33=1* M_2* 50.60\\\\M_2=(1* 0.0944* 38.33)/(1* 50.60)=0.0715M`

Hence, the concentration of original ammonia solution is 0.0715 M