# 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"# tobegin 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** **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 **s of something do NOT change with volume** .

Since the ** equimolar** with 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.