1. 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

2. Note: Salt is not always NaCl.
Write the neutralization reaction when H2SO4 reacts with KOH. Label the acid, the base, and the salt.
H2SO KOH  2 2
H2O
K2SO4
acid
base
salt
Episode 1103

3. 2HNO3 + Mg(OH)2  2H2O + Mg(NO3)2
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

4. 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

5. If neutral, moles of [H+] = [OH-]
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

6. Find the molarity of a sample of HCl by neutralizing it with 0
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

7. 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

8. (#moles H+/1 molea )(Ma)(Va) = (#moles OH-/1 moleb )(Mb)(Vb)
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

9. (#moles H+/1 molea )(Ma)(Va) = (#moles OH-/1 moleb )(Mb)(Vb)
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