Question

100 mL of 1 M HCl is mixed with 50 mL of 2 M HCl. Hence, $[{{H}_{3}}{{O}^{+}}]$is________.
(A) 1.00 M
(B) 1.50 M
(C) 1.33 M
(D) 3.00 M

Answer 1

There is a formula to calculate the concentration of the solution.
$M=\dfrac{n}{V}$
M = molarity of the solution
n = number of moles
V = volume of the solution

Complete step by step solution:
-In the question it is given that 100 mL of the 1 M HCl with 50 mL of 2 M HCl.
-We have to find the concentration of the $[{{H}_{3}}{{O}^{+}}]$ in the resulting solution.
-Number of moles in 100 mL of 1 M HCl is
${{M}_{1}}=\dfrac{{{n}_{1}}}{{{V}_{1}}}$
${{M}_{1}}$ = Molarity of HCl = 1 M
${{n}_{1}}$= number of moles of HCl = ?
${{V}_{1}}$= Volume of the HCl = 100 mL = 0.1 L
Therefore

& {{M}_{1}}=\dfrac{{{n}_{1}}}{{{V}_{1}}} \\\ & {{n}_{1}}={{M}_{1}}\times {{V}_{1}} \\\ & {{n}_{1}}=1\times 0.1 \\\ & {{n}_{1}}=0.1 \\\ \end{aligned}$$ -Therefore $${{n}_{1}}$$= number of moles of HCl = 0.1 -Number of moles in 50 mL of 2 M HCl is $${{M}_{2}}=\dfrac{{{n}_{2}}}{{{V}_{2}}}$$ $${{M}_{2}}$$ = Molarity of HCl = 2 M $${{n}_{2}}$$= number of moles of HCl = ? $${{V}_{2}}$$= Volume of the HCl = 50 mL = 0.05 L $$\begin{aligned} & {{M}_{2}}=\dfrac{{{n}_{2}}}{{{V}_{2}}} \\\ & {{n}_{2}}={{M}_{2}}\times {{V}_{2}} \\\ & {{n}_{2}}=2\times 0.05 \\\ & {{n}_{2}}=0.1 \\\ \end{aligned}$$ -Therefore $${{n}_{2}}$$= number of moles of HCl = 0.1 -Total number of moles of the mixture of 100 mL of 1 M HCl and 50 mL of 2 M HCl is $$\begin{aligned} & n={{n}_{1}}+{{n}_{2}} \\\ & n=0.1+0.1 \\\ & n=0.2 \\\ \end{aligned}$$ -Total volume of the solution by adding 100 mL of 1 M HCl and 50 mL of 2 M HCl is V = 100+50 = 150 mL = 0.15 L -Therefore Total concentration of the solution is $$\begin{aligned} & M=\dfrac{n}{V} \\\ & M=\dfrac{0.2}{0.15}=1.33M \\\ \end{aligned}$$ M = molarity of the solution n = number of moles V = volume of the solution -$$[{{H}_{3}}{{O}^{+}}]$$ = M = 1.33 M. -The molarity or the concentration of $$[{{H}_{3}}{{O}^{+}}]$$ = 1.33 M. **Therefore the correct option is C.** **Note:** We are calculating the total concentration of the resulting solution so we have to consider the number moles of the 100 mL of 1 M HCl and 50 mL of 2 M HCl. Then only we will get the concentration (M) of the solution.