Neutralization Reactions and Titrations

Neutralization Reactions

Neutralization reactions occur when an acid and a base react, forming salt and water.

General Reaction

$Acid + Base \rightarrow Salt + Water$

Double Replacement Examples

• $HCl + NaOH \rightarrow NaCl + H_2O$
• $HNO<em>3 + Mg(OH)</em>2 \rightarrow Mg(NO<em>3)</em>2 + H_2O$
• $H<em>2CO</em>3 + KOH \rightarrow K<em>2CO</em>3 + H_2O$

Stoichiometry with Acids and Bases

The steps are the same as in regular stoichiometry problems:

1. Balance the equation.
2. Convert the given information into moles.
3. Use the mole ratio (coefficients from the balanced equation).
4. Convert the answer to the desired units.

Practice Problems

Problem 1

What is the molarity of a $NaOH$ solution if 80.0 mL are needed to neutralize 10.0 mL of 0.10 M $HCl$?

1. Balanced Equation:
   $HCl + NaOH \rightarrow NaCl + H_2O$
2. Convert given data into moles:
   $0.10 M HCl = \frac{? \text{ moles}}{0.0100 L}$
   $Moles = 0.0010 \text{ moles of } HCl$
3. Use the mole ratio:
   $0.0010 \text{ moles } HCl \times \frac{1 \text{ mol } NaOH}{1 \text{ mol } HCl} = 0.0010 \text{ moles } NaOH$
4. Convert to desired unit (molarity):
   $M = \frac{\text{moles}}{L}$
   $M = \frac{0.0010 \text{ moles}}{0.0800 L}$
   $M = 0.013 M NaOH$

Problem 2

How many mL of 6.00 M $HNO<em>3$ are needed to completely neutralize 2.00 g of $Mg(OH)</em>2$?

1. Balanced Equation:
   $2 HNO<em>3 + Mg(OH)</em>2 \rightarrow Mg(NO<em>3)</em>2 + 2 H_2O$
2. Convert given data into moles:
   $2.00 g Mg(OH)<em>2 \times \frac{1 \text{ mol } Mg(OH)</em>2}{58.33 g} = 0.0343 \text{ moles } Mg(OH)_2$
3. Use the mole ratio:
   $0.0343 \text{ moles } Mg(OH)<em>2 \times \frac{2 \text{ mol } HNO</em>3}{1 \text{ mol } Mg(OH)<em>2} = 0.0686 \text{ moles } HNO</em>3$
4. Convert to desired unit (volume):
   $M = \frac{\text{moles}}{L}$
   $6.00 M = \frac{0.0686 \text{ moles}}{? L}$
   $L = 0.0114 L$
   $0.0114 L \times \frac{1000 mL}{1 L} = 11.4 \text{ mL of the acid}$

Determining Neutralization

Acid-base reactions are usually colorless, so you need a way to determine when neutralization has occurred. Options include:

• Indicator: A substance that changes color when the endpoint of the reaction is reached.
• pH meter: Measures the pH of the solution.

Common Indicators

Titrations

A titration is a laboratory procedure that uses a neutralization reaction to determine the concentration of an unknown acid or base.

• A solution of known concentration (a standard) is reacted with the unknown solution, with careful volume measurements.

Equivalence and Endpoint

• Equivalence Point: The point in the titration when equal moles of acid and base have reacted (neutralization has occurred).
• Endpoint: The point at which the titration is stopped, usually signaled by a color change of an indicator or a pH meter reading.

Sample Data Problem

20.0 mL of a 0.10 M $HCl$ solution (standard) is added to a clean flask. 2 drops of phenolphthalein indicator are added. A buret is filled with the unknown $NaOH$ solution. Base is added to the flask until the color turns pink, which occurs when 8.00 mL of $NaOH$ has been added. What is the concentration of the base ($NaOH$)?

Solution

1. Balanced Equation:
   $HCl + NaOH \rightarrow NaCl + H_2O$
2. Convert given data into moles:
   $0.10 M HCl = \frac{? \text{ moles}}{0.0200 L}$
   $Moles = 0.0020 \text{ moles of } HCl$
3. Use the mole ratio:
   $0.0020 \text{ moles } HCl \times \frac{1 \text{ mol } NaOH}{1 \text{ mol } HCl} = 0.0020 \text{ moles } NaOH$
4. Convert to desired unit (molarity):
   $M = \frac{\text{moles}}{L}$
   $M = \frac{0.0020 \text{ moles}}{0.0080 L}$
   $M = 0.25 M NaOH$

Practice Problem

What is the molarity of a $KOH$ solution if 40.0 mL are needed to neutralize 20.0 mL of 0.10 M $H<em>2SO</em>4$?

1. Balanced Equation:
   $H<em>2SO</em>4 + 2 KOH \rightarrow K<em>2SO</em>4 + 2 H_2O$
2. Convert given data into moles:
   $0.10 M H<em>2SO</em>4 = \frac{? \text{ moles}}{0.0200 L}$
   $Moles = 0.0020 \text{ moles of } H<em>2SO</em>4$
3. Use the mole ratio:
   $0.0020 \text{ moles } H<em>2SO</em>4 \times \frac{2 \text{ mol } KOH}{1 \text{ mol } H<em>2SO</em>4} = 0.0040 \text{ moles } KOH$
4. Convert to desired unit (molarity):
   $M = \frac{\text{moles}}{L}$
   $M = \frac{0.0040 \text{ moles}}{0.0400 L}$
   $M = 0.10 M KOH$