# Question 7a6fe

Dec 26, 2015

$\text{0.01 N}$

#### Explanation:

A short answer - the normality of the final solution will be $\text{0.01 N}$, since the volume of this solution is five times larger than the volume of the initial solution.

Now for the long answer. As you know, normality is defined as number of equivalents per liter of solution.

In the context of acid - base chemistry, an equivalent is simply a mole of hydronium ions, ${\text{H"_3"O}}^{+}$.

In the case of hydrochloric acid, $\text{HCl}$, which is a strong acid that dissociates completely in aqueous solution, you can say that every mole of acid will produce one mole of hydronium ions in solution

${\text{HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

This means that for hydrochloric acid, molarity is equivalent to normality.

So, you're dealing with a $\text{0.05-N}$ hydrochloric acid, $\text{HCl}$, solution that has an initial volume of $\text{250 mL}$. The new volume of the solution upon the addition of the water will be

${V}_{\text{sol" = "250 mL" + "1000 mL" = "1250 mL}}$

As you can see, adding that much water to the starting solution is equivalent to diluting it by a factor of $5$, since

"D.F." = (1250 color(red)(cancel(color(black)("mL"))))/(250color(red)(cancel(color(black)("mL")))) = 5

Here $\text{D.F.}$, which is equal to the final volume of the solution divided by the nitial volume of the sample, represents the dilution factor.

Since normality is equal to

$\textcolor{b l u e}{\text{normality" = "equivalents"/"liters of solution}}$

you can say that your initial solution contained

$0.05 \text{eq."/color(red)(cancel(color(black)("L"))) * 250 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.0125 eq.}$

Since the volume of the final solution is now equal to $\text{1250 mL}$, its normality will be equal to

"0.0125 eq."/(1250 * 10^(-3)"L") = "0.01 eq./L" = color(green)("0.01 N")

Alternatively, you can use the formula for dilution calculations

$\textcolor{b l u e}{{\text{normality"_1 xx V_1 = "normality}}_{2} \times {V}_{2}}$

This will once again get you

${\text{normality"_2 = V_1/V_2 xx "normality}}_{1}$

"normality"_2 = (250 color(red)(cancel(color(black)("mL"))))/(1250color(red)(cancel(color(black)("mL")))) * "0.05 N" = color(green)("0.01 N")#