salt

Report
What type of reaction?
HCl + NaOH  H2O + NaCl
• A neutralization reaction is the reaction
between an acid and a base to produce a
salt plus water.
• A salt is any compound containing the
cation of a base and an anion from an
acid.
– NaOH (base) -- Na+ cation
– HCL (acid) -- Cl- anion
– SALT -- NaCl
Episode 1103
Note: Salt is not always NaCl.
• Write the neutralization reaction when
H2SO4 reacts with KOH. Label the acid,
the base, and the salt.
H2SO4 + 2 KOH  2 H2O + K2SO4
acid
base
salt
Episode 1103
Write the neutralization reaction when nitric
acid reacts with magnesium hydroxide.
• Nitric acid
– HNO3
• Magnesium hydroxide
– Mg(OH)2
You know the products is going to be water
and a salt
• Formulate the salt
• Write the equation and balance.
2HNO3 + Mg(OH)2  2H2O + Mg(NO3)2
Episode 1103
Titrations
• A titration is a
laboratory method used
to determine the
concentration of an acid
or base in solution by
performing a
neutralization reaction
with a standard solution.
• In a neutral solution, the
moles of hydrogen ions
must be equal to the
moles of hydroxide ions.
Episode 1103
• To find moles of [H+]
– Moles H+ = (#moles H+/1 moleacid )(Macid)(Vacid)
• To find moles of [OH-]
– Moles OH- = (#moles OH-/1 molebase )(Mbase)(Vbase)
• If neutral, moles of [H+] = [OH-]
– (#moles H+/1 moleacid )(Macid)(Vacid) = (#moles OH-/1
molebase )(Mbase)(Vbase)
Episode 1103
Find the molarity of a sample of HCl
by neutralizing it with 0.5 M NaOH.
– Record volume of HCl solution.
– Add indicator to HCl solution
– Slowly add NaOH solution from burrette.
• The endpoint of a titration is the point
at which the indicator changes color
indicating the neutralization has been
reached so the moles of hydrogen ions
and moles of hydroxide ions are equal.
– Record volume of NaOH solution added.
Episode 1103
Titration Calculation
(#moles H+/1 molea )(Ma)(Va) = (#moles OH-/1 moleb )(Mb)(Vb)
(1mol H+/1 mol HCl )(Ma)(50.0 mL) = (1 mol OH-/1 molNaOH )(0.5M)(20.2 ml)
(Ma)(50.0 mL) = (0.5M)(20.2 ml)
Ma = (0.5 M)(20.2 mL)/50.0 mL
Ma= 0.20 M HCl
Episode 1103
In a titration of HCl and KOH, 50.0 mL of the
base were required to neutralize 10.0 mL of a
3.0 M HCl. What is the molarity of the KOH?
(#moles H+/1 molea )(Ma)(Va) = (#moles OH-/1 moleb )(Mb)(Vb)
(1mol H+/1 mol HCl )(3.0M)(10.0 mL) = (1 mol OH-/1 molNaOH )(Mb)(50.0 ml)
(3.0M)(10.0 mL) = (Mb)(50.0 ml)
Mb = (3.0M)(10.0 mL)/50.0 mL
Mb= 0.60 M KOH
Episode 1103
60.0 mL of 0.50 molar NaOH were needed to
neutralize 30.0 mL of H2SO4. What is the
molarity of the acid?
(#moles H+/1 molea )(Ma)(Va) = (#moles OH-/1 moleb )(Mb)(Vb)
(2mol H+/1 mol HCl )(Ma)(30.0 mL) = (1 mol OH-/1 molNaOH )(0.5M)(60.0 ml)
2(Ma)(30.0 mL) = (0.5M)(60.0 ml)
Ma = (0.5 M)(60.0 mL)/(2)(30.0 mL)
Ma= 0.50 M H2SO4
Episode 1103

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