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nalin [4]
3 years ago
14

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

ia solution?
Chemistry
1 answer:
zzz [600]3 years ago
4 0

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?

<u>Answer:</u> The concentration of original ammonia solution is 0.0715 M

<u>Explanation:</u>

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

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