Chemistry lab question: Part C: Using table 1, identify the unknown base *l have included part a and b, which includes the pOH and Kb. I will also attach table 1.
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Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.
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Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.
Chemistry lab question:
Part C: Using table 1, identify the unknown base
*l have included part a and b, which includes the pOH and Kb. I will also attach table 1.
![### Chemistry Titration Problem
#### Problem Statement:
1. **Unknown base solution** = 10ml
#### (a) Titration with Hydrochloric Acid:
- **Titrated with known concentration** of HCl of molar volume = 0.100M.
- **Given graph equivalence point** = pH 3.92.
- To find:
- **Kb** = ?
- **pOH** = ?
**Half Equivalence Point:**
- At the half equivalence point, pH = pKa.
- Relationship:
\[ \text{pKa} + \text{pKb} = 14 \]
Given:
\[ \text{pKa} = 3.92 \]
\[ \text{pKb} = 14 - 3.92 = 10.08 \]
Therefore:
\[ \text{pOH} = 10.08 \]
Subsequently, the OH⁻ concentration is:
\[ \text{[OH]⁻} = 10^{-10.08} \]
#### (b) Dissociation of NaOH:
The dissociation equation of NaOH is:
\[ \text{NaOH} \leftrightarrow \text{Na}^+ + \text{OH}^- \]
Initial concentrations:
\[ \text{NaOH} = 0.100 \,M \]
\[ \text{Na}^+ = 0 \]
\[ \text{OH}^- = 0 \]
Changes in concentrations:
\[ \text{NaOH}: (0.100 - x) \]
\[ \text{Na}^+: x \]
\[ \text{OH}^-: x \]
At equilibrium:
\[ \text{[OH]⁻} = x = 10^{-10.08} \]
The base dissociation constant (Kb) is given by:
\[ K_b = \frac{x^2}{0.100 - x} \]
Substituting values:
\[ K_b = \frac{(10^{-10.08})^2}{0.100 - 10^{-10.08}} \]
Performing the calculation:
\[ K_b \approx 1.6 \times 10^{-4} \]
#### (c)
*Note: The third part of the problem is not included in the provided text.*
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