Question

If 17.4 mL of 0.800 M HCl solution are needed to neutralize 5.00 mL of a household ammonia solution, what is the molar concentration of the ammonia?

Answer 1

NH3  +HCl  ---->  NH4Cl

moles   of  HCl  used =    (0.8  x  17.4) /1000=  0.0139 moles
by  use   of reacting  ratio  between  HCl  to   NH4Cl    which is 1:1    therefore  the  moles   of  NH4Cl  is  also  =  0.0139 moles
molar   concentration  =  moles  /volume  in  liters 

molar  concentration  is  therefore=    (0.0139/5)  x1000 =  2.7 M

Answer 2

Answer: The concentration of ammonia is 2.784 M

Explanation:

To calculate the concentration of acid, 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 ammonia

We are given:

$n_1=1\\M_1=0.800M\\V_1=17.4mL\\n_2=1\\M_2=?M\\V_2=5.00mL$

Putting values in above equation, we get:

$1\times 0.800\times 17.4=1\times M_2\times 5.00\\\\M_2=2.784M$

Hence, the concentration of ammonia is 2.784 M