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?

1 Answer
Truong-Son N.
Jul 3, 2016

I got #"0.01 L"#, or #"10 cm"^3#.

CLEARING UP UNITS

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

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

So, let's just say we have #"0.2 M"# #"HCl"#, #"0.1 M"# #"NaOH"#, and #"0.5 M"# #"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 #"H"^(+)# was in solution before we did anything.
  2. Find out how much #"H"^(+)# is now in solution after adding #"50 cm"^3# #"NaOH"#.
  3. Find out how much #"KOH"# needs to be added to neutralize the remaining #"H"^(+)# in solution.

STARTING CONCENTRATION OF PROTONS

The volume #V_"HCl"# of #"0.2 M"# #"HCl"# is:

#V_"HCl" = "50 cm"^3 = "50 mL" = "0.050 L"#

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

#color(green)(n_"HCl") = "0.2 M"# #xx# #"0.050 L"#

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

CURRENT CONCENTRATION OF PROTONS

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

#n_"NaOH" = "0.1 M"# #xx# #"0.050 L"# #=# #"0.005 mols"# of #"NaOH",#

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

Since the #"OH"^(-)# produced by dissociating #"KOH"# is equimolar with the #"H"^(+)# produced by dissociating #"HCl"#, that means we need #"0.005 mols"# #"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.

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