Question

If 18.8 mLmL of 0.800 M HClM HCl solution are needed to neutralize 5.00 mLmL of a household ammonia solution, what is the molar concentration of the ammonia? Express your answer with the appropriate units.

Answer 1

Answer: The molar concentration of ammonia is 3.008 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:

$n_1=1\\M_1=0.800M\\V_1=18.8mL\\n_2=1\\M_2=?M\\V_2=5.00mL$

Putting values in above equation, we get:

$1\times 0.800\times 18.8=1\times M_2\times 5.00\\\\x=\frac{1\times 0.800\times 18.8}{1\times 5.00}=3.008M$

Hence, the molar concentration of ammonia is 3.008 M