Certainly! To determine the normality of the solution as a base, we need to follow several steps. Normality is a measure of concentration equivalent to molarity multiplied by the number of equivalents. In the case of bases, this typically refers to the number of hydroxide [tex]\( (OH^-) \)[/tex] ions supplied by the compound.
Here are the steps to solve this problem:
### 1. Determine the molar masses of NaOH and Ba(OH)[tex]\(_2\)[/tex]
- NaOH:
- Sodium (Na): 22.99 g/mol
- Oxygen (O): 16.00 g/mol
- Hydrogen (H): 1.01 g/mol
- Molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
- Ba(OH)[tex]\(_2\)[/tex]:
- Barium (Ba): 137.33 g/mol
- Oxygen (O): 16.00 g/mol x 2 = 32.00 g/mol
- Hydrogen (H): 1.01 g/mol x 2 = 2.02 g/mol
- Molar mass of Ba(OH)[tex]\(_2\)[/tex] = 137.33 + 32.00 + 2.02 = 171.35 g/mol
### 2. Calculate the moles of NaOH and Ba(OH)[tex]\(_2\)[/tex]
- Moles of NaOH:
[tex]\[
\text{moles of NaOH} = \frac{7.82 \text{ g}}{40.00 \text{ g/mol}} = 0.1955 \text{ moles}
\][/tex]
- Moles of Ba(OH)[tex]\(_2\)[/tex]:
[tex]\[
\text{moles of Ba(OH)}_2 = \frac{9.26 \text{ g}}{171.35 \text{ g/mol}} = 0.05405 \text{ moles}
\][/tex]
### 3. Determine the number of equivalents for each base
- NaOH:
Each NaOH molecule provides 1 hydroxide ion (OH[tex]\(^-\)[/tex]), so the number of equivalents is the same as the number of moles:
[tex]\[
\text{equivalents of NaOH} = 0.1955 \text{ equivalents}
\][/tex]
- Ba(OH)[tex]\(_2\)[/tex]:
Each Ba(OH)[tex]\(_2\)[/tex] molecule provides 2 hydroxide ions (OH[tex]\(^-\)[/tex]):
[tex]\[
\text{equivalents of Ba(OH)}_2 = 0.05405 \times 2 = 0.1081 \text{ equivalents}
\][/tex]
### 4. Calculate the total number of equivalents in the solution
[tex]\[
\text{total equivalents} = 0.1955 + 0.1081 = 0.3036 \text{ equivalents}
\][/tex]
### 5. Determine the normality of the solution
Normality (N) is defined as the number of equivalents per liter of solution. The total volume of the solution is 500 mL, which is equivalent to 0.5 L.
[tex]\[
\text{Normality} = \frac{\text{total equivalents}}{\text{volume in liters}} = \frac{0.3036 \text{ equivalents}}{0.5 \text{ L}} = 0.6072 \text{ N}
\][/tex]
### Conclusion
The normality of the solution as a base is [tex]\( 0.6072 \text{ N} \)[/tex].
