LS 3 Exercise Solutions: General and Inorganic Chemistry - pH Calculations and Salt Hydrolysis

Institution: Institut für Chemie und Biochemie – Anorganische Chemie (Institute for Chemistry and Biochemistry – Inorganic Chemistry).
Course: Übung zur Vorlesung „Allgemeine und Anorganische Chemie“ (Exercise for the Lecture "General and Inorganic Chemistry").
Document Content: Lösung 3. Übungsbogen (Solution to Exercise Sheet 3).

Definition: The keyword provided is the "Nivellierender Effekt des Wassers" (Leveling effect of water).
Description: In aqueous solutions, water exerts a leveling effect on strong acids and bases. This means that no acid stronger than the hydronium ion ( $H_3O^+$ ) and no base stronger than the hydroxide ion ( $OH^-$ ) can exist in water. Any stronger acid will react quantitatively with water to form $H_3O^+$ , and any stronger base will react to form $OH^-$ , thus "leveling" their strength to that of the respective water-derived ions.

The transcript details a calculation involving mass fractions, density, and molar mass to determine a percentage value.
Formula Structure: The calculation is depicted as $\frac{m_{solution} \times \text{percentage factor}}{M \times V} = c$ .
Numerical Execution:
- The specific values utilized in the solution are: $\frac{45}{1410}$ .
- Final Result: The calculated value is exactly $31.91\,\%$ .

Strong Acids: The pH is determined by the negative decadic logarithm of the initial acid concentration ( $c_0$ ).
- Formula: $pH = -\lg(c_0)$
Strong Bases: The pOH is determined by the negative decadic logarithm of the initial base concentration, which is then converted to pH.
- Formula: $pOH = -\lg(c_0)$
- Conversion: $pH = 14 - pOH$
Weak Acids: The pH calculation accounts for the acid dissociation constant ( $pK_a$ ) and the initial concentration.
- Formula: $pH = \frac{1}{2} (pK_a - \lg(c_0))$
Weak Bases: The pOH calculation accounts for the base dissociation constant ( $pK_b$ ) and the initial concentration.
- Formula: $pOH = \frac{1}{2} (pK_b - \lg(c_0))$

Case (a): Hydrochloric Acid ( $HCl$ )
- Classification: Strong Acid.
- Concentration: $0.1\,mol/dm^3$ .
- Calculation: $pH = -\lg(0.1) = -\lg(10^{-1}) = 1$ .
Case (b): Sulfuric Acid ( $H_2SO_4$ )
- Classification: Strong Acid (considering full dissociation).
- Concentration: $0.05\,mol/dm^3$ .
- Calculation Note: Since sulfuric acid provides two protons, the concentration is doubled: $2 \times 0.05 = 0.1$ .
- Calculation: $pH = -\lg(2 \times 0.05) = -\lg(10^{-1}) = 1$ .
Case (c): Acetic Acid ( $HOAc$ )
- Classification: Weak Acid.
- Concentration: $0.1\,mol/dm^3$ .
- Constant: The $pK_a$ for acetic acid is given as $4.75$ .
- Calculation: $pH = \frac{1}{2} (4.75 - \lg(0.1)) = \frac{1}{2} (4.75 - (-1)) = \frac{1}{2} (5.75) = 2.88$ .

Sodium Bromide ( $NaBr$ ): Neutral Solution
- Dissociation Reaction: $NaBr + H_2O \rightarrow NaOH + HBr$ .
- Reasoning: Because it is the salt of a strong base ( $NaOH$ ) and a strong acid ( $HBr$ ), the solution remains neutral.
Ammonium Acetate ( $NH_4OAc$ ): Neutral Solution
- Classification: This is a salt for a combination of a weak acid and a weak base.
- Methodology for pH Determination: The pH is calculated as the average of the $pK_a$ values of the corresponding acid and the conjugate acid of the base.
- Values: The $pK_a$ of the ammonium ion ( $NH_4^+$ ) is $9.25$ ; the $pK_a$ of acetic acid ( $HOAc$ ) is $4.75$ .
- Calculation: $pH = \frac{1}{2} (pK_a(NH_4^+) + pK_a(HOAc)) = \frac{1}{2} (9.25 + 4.75) = \frac{1}{2} (14) = 7$ .
- Outcome: The resulting solution is neutral.
General Characterizations from Exercise 4:
- Case (b): acidic (sauer).
- Case (c): basic (basisch).
- Case (d): basic (basisch).
- Case (f): results in a $pH = 6.2$ , which is acidic (sauer).

Exercise 6 (b): A specific note indicates that the chemical species in this question reacts acidically (reagiert sauer).