# Suppose that to "50 cm"^3 of "0.2 N HCl" was added "50 cm"^3 of "0.1 N NaOH". What volume of "0.5 N KOH" should be added to neutralize the "HCl" completely?

Jul 3, 2016

I got $\text{0.01 L}$, or ${\text{10 cm}}^{3}$.

CLEARING UP UNITS

First off, the unit of normality ($\text{N}$) means that the concentration is described based on the number of ions that dissociate, so for example, ${\text{H"_2"SO}}_{4}$ is $\text{2 N}$ in ${\text{H}}^{+}$ but $\text{1 N}$ in ${\text{SO}}_{4}^{2 -}$.

For $\text{HCl}$, $\text{NaOH}$, and $\text{KOH}$, it is the same as saying molarity ($\text{M}$) in this case, because there is only one ${\text{H}}^{+}$ and one ${\text{OH}}^{-}$ going into solution for each $\text{HCl}$, $\text{NaOH}$, or $\text{KOH}$ dissociating (with respect to their corresponding ions).

So, let's just say we have $\text{0.2 M}$ $\text{HCl}$, $\text{0.1 M}$ $\text{NaOH}$, and $\text{0.5 M}$ $\text{KOH}$. It's the same thing in this case, anyway, and more familiar to us.

After clearing up the units, this problem is basically a segmented equimolar neutralization of acid. All we have to do is:

1. Find out how much ${\text{H}}^{+}$ was in solution before we did anything.
2. Find out how much ${\text{H}}^{+}$ is now in solution after adding ${\text{50 cm}}^{3}$ $\text{NaOH}$.
3. Find out how much $\text{KOH}$ needs to be added to neutralize the remaining ${\text{H}}^{+}$ in solution.

STARTING CONCENTRATION OF PROTONS

The volume ${V}_{\text{HCl}}$ of $\text{0.2 M}$ $\text{HCl}$ is:

${V}_{\text{HCl" = "50 cm"^3 = "50 mL" = "0.050 L}}$

So the $\text{mol}$s of $\text{HCl}$ that dissociated in water to give ${\text{H}}^{+}$ is:

color(green)(n_"HCl") = "0.2 M" $\times$ $\text{0.050 L}$

$=$ $\textcolor{g r e e n}{\text{0.01 mols}}$ of $\text{HCl}$ to begin with.

CURRENT CONCENTRATION OF PROTONS

We know that ${\text{50 cm}}^{3}$ of $\text{NaOH}$ was added, so by adding $\text{0.050 L}$ of $\text{NaOH}$, we have added

${n}_{\text{NaOH" = "0.1 M}}$ $\times$ $\text{0.050 L}$ $=$ $\text{0.005 mols}$ of $\text{NaOH} ,$

meaning that half of the $\setminus m a t h b f \left(\text{HCl}\right)$ has been neutralized; we went from

n_("HCl","starting") = "0.01 mols"

to

color(green)(n_("HCl","current")) = 0.01 - "0.005 mols" = color(green)("0.005 mols HCl").

HOW MUCH KOH IS NEEDED

Therefore, when we add $\text{0.5 M}$ $\text{KOH}$, we just need to neutralize $\text{0.005 mols HCl}$, irrespective of the solution's current volume---the $\setminus m a t h b f \left(\text{mol}\right)$s of something do NOT change with volume .

Since the ${\text{OH}}^{-}$ produced by dissociating $\text{KOH}$ is equimolar with the ${\text{H}}^{+}$ produced by dissociating $\text{HCl}$, that means we need $\text{0.005 mols}$ $\text{KOH}$. Therefore, we need this volume:

color(blue)(V_"KOH") = 0.005 cancel("mols KOH") xx "1 L"/(0.5 cancel("mols KOH"))

$=$ $\textcolor{b l u e}{\text{0.01 L KOH}}$

I looked up the answer online and found this. Since the actual answer was ${\text{10 cm}}^{3}$...

"10 cm"^3 = "10 mL" = color(blue)("0.01 L")

$\therefore$ our answer is correct.