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

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.

Answer 2

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 $NH_3$.

$(0.293M)\times (0.0273L)=M_2\times (0.0426L)$

$M_2=0.1877M=0.188M$

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