Question

If 27.3 mL of 0.293 M HBr solution completely neutralizes 42.6 mL of NH3, what is the concentration of the NH3 solution? NH3 + HBr yields NH4Br 0.188 M 0.387 M 0.586 M 2.18 M

Answer 1

Answer : The correct option is, 0.188 M

Solution : Given,

Volume of HBr solution = 27.3 ml = 0.0273 L (1 L = 1000 ml)

Molarity of HBr solution = 0.293 M

Volume of
`NH_3` solution = 42.6 ml = 0.0426 L

The complete neutralization reaction is,

`NH_3+HBr\rightarrow NH_4Br`

Formula used :

`M_1V_1=M_2V_2`

where,

`M_1` = Molarity of HBr solution

`V_1` = volume of HBr solution

`M_2` = Molarity of
`NH_3` solution

`V_2` = volume of
`NH_3` solution

Now put all the given values in the above formula, we get the concentration of
`NH_3`.

`(0.293M)* (0.0273L)=M_2* (0.0426L)`

`M_2=0.1877M=0.188M`

Therefore, the concentration of the
`NH_3` is, 0.188 M

Answer 2

According to reaction equation: NH3+HBr=NH4Br, 1 mole of NH3 is required to react with 1 mole of HBr.  HBr solution contains 0.0273L * 0.293M = 0.008 moles of HBr.  Therefore 0.008 mole of NH3 is reacted. So the concentration of NH3 is 0.008 mole/0.0426 L = 0.188M.  The answer is 0.188M.