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

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:

$1\times 0.0944\times 38.33=1\times M_2\times 50.60\\\\M_2=\frac{1\times 0.0944\times 38.33}{1\times 50.60}=0.0715M$

Hence, the concentration of original ammonia solution is 0.0715 M