Answer
$c(NaCl) = 1.1578 M$
Work Step by Step
First of all, we can calculate the amount of hydrochloric acid and sodium - hydroxide present in their solutions:
$n(HCl) = c(HCl) \times V(HCl) = 3.35 \frac{mol}{dm^3} \times 0.225 dm^3 = 0.75375 mol$
$n(NaOH) = c(NaOH) \times V(NaOH) = 1.77 \frac{mol}{dm^3} \times 0.426 dm^3 = 0.75402 mol$
$NaOH$ and $HCl$ react according to the following equation:
$NaOH + HCl \rightarrow NaCl + H_{2}O$
Since the stoichiometric ratio of $NaOH$ and $HCl$ is equal to $1:1$, the limiting reagent in this situation is hydrochloric acid, because $n(NaOH) > n(HCl)$. Therefore, the amount of sodium - chloride obtained is equal to the amount of hydrochloric acid (according to the coefficients in the equation), $n(NaCl) = 0.75375 mol$.
Final volume of the solution is equal to $V_{fin} = V(HCl) + V(NaOH) = 0.225 dm^3 + 0.426 dm^3 = 0.651 dm^3$
Finally, $c(NaCl) = \frac{n(NaCl)}{V_{fin}} = \frac{0.75375 mol}{0.651dm^3} = 1.1578M$
