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

Question

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

1 Answer

Impact of this question

5331 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-03T03:25:33

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

CLEARING UP UNITS

First off, the unit of normality ( $N$ $"N"$ ) means that the concentration is described based on the number of ions that dissociate, so for example, $H_{2} {SO}_{4}$ $"H"_2"SO"_4$ is $2 N$ $"2 N"$ in $H^{+}$ $"H"^(+)$ but $1 N$ $"1 N"$ in ${SO}_{4}^{2 -}$ $"SO"_4^(2-)$ .

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

So, let's just say we have $0.2 M$ $"0.2 M"$ $HCl$ $"HCl"$ , $0.1 M$ $"0.1 M"$ $NaOH$ $"NaOH"$ , and $0.5 M$ $"0.5 M"$ $KOH$ $"KOH"$ . 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:

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

STARTING CONCENTRATION OF PROTONS

The volume $V_{HCl}$ $V_"HCl"$ of $0.2 M$ $"0.2 M"$ $HCl$ $"HCl"$ is:

$V_{HCl} = {50 cm}^{3} = 50 mL = 0.050 L$ $V_"HCl" = "50 cm"^3 = "50 mL" = "0.050 L"$

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

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

$=$ $=$ $0.01 mols$ $color(green)("0.01 mols")$ of $HCl$ $"HCl"$ to begin with.

CURRENT CONCENTRATION OF PROTONS

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

$n_{NaOH} = 0.1 M$ $n_"NaOH" = "0.1 M"$ $\times$ $xx$ $0.050 L$ $"0.050 L"$ $=$ $=$ $0.005 mols$ $"0.005 mols"$ of $NaOH,$ $"NaOH",$

meaning that half of the $HCl$ $\mathbf("HCl")$ has been neutralized; we went from

$n_{HCl, starting} = 0.01 mols$ $n_("HCl","starting") = "0.01 mols"$

to

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

HOW MUCH KOH IS NEEDED

Therefore, when we add $0.5 M$ $"0.5 M"$ $KOH$ $"KOH"$ , we just need to neutralize $0.005 mols HCl$ $"0.005 mols HCl"$ , irrespective of the solution's current volume---the $mol$ $\mathbf("mol")$ s of something do NOT change with volume .

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

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

$=$ $color(blue)("0.01 L KOH")$

I looked up the answer online and found this. Since the actual answer was $"10 cm"^3$ ...

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

$:.$ our answer is correct.

Science

Math

Humanities

... and beyond

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

1 Answer

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

Suppose that to "50 cm"^350 cm3"50 cm"^3 of "0.2 N HCl"0.2 N HCl"0.2 N HCl" was added "50 cm"^350 cm3"50 cm"^3 of "0.1 N NaOH"0.1 N NaOH"0.1 N NaOH". What volume of "0.5 N KOH"0.5 N KOH"0.5 N KOH" should be added to neutralize the "HCl"HCl"HCl" completely?

1 Answer

Related questions

Impact of this question

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