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
I got
CLEARING UP UNITS
First off, the unit of normality (
For
So, let's just say we have
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"^(+)# was in solution before we did anything. - Find out how much
#"H"^(+)# is now in solution after adding#"50 cm"^3# #"NaOH"# . - Find out how much
#"KOH"# needs to be added to neutralize the remaining#"H"^(+)# in solution.
STARTING CONCENTRATION OF PROTONS
The volume
#V_"HCl" = "50 cm"^3 = "50 mL" = "0.050 L"#
So the
#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
#n_"NaOH" = "0.1 M"# #xx# #"0.050 L"# #=# #"0.005 mols"# of#"NaOH",#
meaning that half of the
#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
Since the
#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 mL" = color(blue)("0.01 L")#
#:.# our answer is correct.