# Question #7a6fe

##### 1 Answer

#### Explanation:

A *short answer* - the normality of the final solution will be **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,

In the case of hydrochloric acid, **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

#"HCl"_text((aq]) + "H"_2"O"_text((l]) -> "H"_3"O"_text((aq])^(+) + "Cl"_text((aq])^(-)#

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

So, you're dealing with a

#V_"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

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

Here *final volume* of the solution divided by the *nitial volume* of the sample, represents the dilution factor.

Since normality is equal to

#color(blue)("normality" = "equivalents"/"liters of solution")#

you can say that your initial solution contained

#0.05"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

#"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

#color(blue)("normality"_1 xx V_1 = "normality"_2 xx V_2)#

This will once again get you

#"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")#