1. Neutralization

2. Acid + Base  Salt + Water
Neutralization
Acid + Base  Salt + Water
Water is formed when the H+ ion from the acid contacts the OH- ion from the base.
H+ + OH-  H2O
Salt is an ionic compound formed from the other parts of the acid and base.
ex: NaCl, KF, Li2SO4

3. Terms
Titration is a lab technique used to determine the concentration of an acid or base.
Equivalence point is when the number of moles of hydrogen ions equals the number of moles of hydroxide ions.
This does NOT necessarily mean pH is 7, as the salt may change the pH of the solution
End point is when the indicator changes colors.
(Meaning the solution has changed from acidic to basic or vise versa)

4. Indicators
An indicator is a chemical that changes color at a specific pH.
Many indicators are a mixture of many chemicals. Some are found naturally
Cabbage Juice
Universal Indicator

6. Steps for Titrations
A measured volume of an acid (or base) solution of an unknown concentration is placed in a flask
Several drops of indicator are added to the flask
Measured volumes of a base (or acid, if a base is in the flask) of known concentration (standard solution) are placed in a buret and slowly added into the acid until the indicator changes color.

7. Lab Example
Phenolphthalein indicator is clear in acid and pink in base.
The endpoint is when 1 extra drop of base is added after equivalence is reached, changing the indicator color.

8. Titration The concentration of an unknown solution can be determined
Titrant (unknown solution) added from a buret
Indicators are chemicals added to help determine when a reaction is complete
The endpoint of the titration occurs when the reaction is complete
This is also known as the equivalence point – the point in the titration when the number of moles of OH– added equal the number of moles of H+ originally in solution

9. Plotting a Titration Curve
If the pH is measured throughout titration, this plot can be generated.

10. Neutralization Calculations
If the number of moles of acid = moles of base, you can use stoichiometry to find the Molarity of the unknown
Example - A 43.0 mL volume of NaOH was titrated with 32.0 mL of M HCl. What is the molarity of the sodium hydroxide solution?

11. Step 1: how many moles of HCl were needed?
L (0.100 moles/L) = moles of HCl
Step 2: How many moles of H are in each mole of the acid?
HCl is monoprotic, so 1mole HCl:1 mole H, meaning moles of H+
Step 3: Moles H+ = Moles OH-
so there must have been moles of OH in the solution

12. Step 5: calculate Molarity
Step 4: How many moles of OH are in each mole of base?
1mole OH/ 1 mole NaOH, so moles NaOH
Step 5: calculate Molarity
moles NaOH
0.0430L of solution
NaOH = M

13. Example
A volume of 50.0 mL of 1.20 M H3PO4 neutralizes 45.5 mL of a Ca(OH)2 solution. What is the molarity of the Ca(OH)2 solution?

14. Neutralization Calculations (quicker)
na(Mb) (Vb) = nb(Ma) (Va)
The same formula as dilutions.
MaVa are for the acidic solution and Mb Vb are for the basic solution
n are the coefficients from the balanced equation. (note that they are backwards)
*You must have a properly balanced equation for this to work!

15. na(Mb) (Vb) = nb(Ma) (Va)
Example
Solutions of sodium hydroxide (NaOH) are used to unclog drains. A 43.0 mL volume of NaOH was titrated with 32.0 mL of M HCl. What is the molarity of the sodium hydroxide solution?
NaOH + HCl  NaCl + H2O
na(Mb) (Vb) = nb(Ma) (Va)

16. na(Mb) (Vb) = nb(Ma) (Va)
Sample Problem
A volume of 25. mL of M H2SO4 neutralizes 40. mL of a NaOH solution. What is the concentration of the NaOH solution?
na(Mb) (Vb) = nb(Ma) (Va)