chem frq

Part 1: The Net Ionic Equation (NIE)

The Task: Write the net ionic equation for the reaction between $H N O_{2} (a q)$ and $N a O H (a q)$ .

The Solution:

$H N O_{2} (a q) + O H^{-} (a q) \to N O_{2}^{-} (a q) + H_{2} O (l)$

Why this matters for your test:

Keep Weak Acids Together: Notice $H N O_{2}$ is not written as $H^{+} + N O_{2}^{-}$ . In an NIE, you only split strong electrolytes.
Drop the Spectators: $N a^{+}$ is a spectator ion. If you include it, you lose the point.
The Reaction: A strong base “strips” the proton off the weak acid completely.

Part 2: Initial pH (The ICE Table)

The Task: Calculate the pH of the initial $0.100 M H N O_{2}$ solution ( $K_{a} = 4.0 \times 1 0^{- 4}$ ).

The Math:

Set up the expression: $K_{a} = \frac{[ H ^{+} ] [ N O _{2}^{-} ]}{[ H N O _{2} ]}$
Substitute “x”: $4.0 \times 1 0^{- 4} = \frac{x ^{2}}{0.100 - x} \approx \frac{x ^{2}}{0.100}$
Solve for x: $x = (4.0 \times 1 0^{- 4}) (0.100) = 0.00632 M$
Convert to pH: $p H = - lo g (0.00632) = 2.20$

Part 3: The Particulate Diagram

The Task: A diagram shows 4 molecules of $H N O_{2}$ . If you add 2 units of $N a O H$ , what does the final box look like?

The Logic (Stoichiometry):

You started with 4 $H N O_{2}$ .
You added 2 $O H^{-}$ .
The $O H^{-}$ reacts with 2 of the $H N O_{2}$ molecules to create 2 $N O_{2}^{-}$ ions and **2 $H_{2} O$ **molecules.
Final Result: Your box should contain 2 $H N O_{2}$ (leftover) and 2 $N O_{2}^{-}$ (produced).

Key Takeaway: This is a Buffer because you have equal amounts of the weak acid and its conjugate base. In this specific box, $p H = p K_{a}$ .

Part 4: Buffer Capacity & Common Ion Effect

The Task: If you add $N a N O_{2} (s)$ to the $H N O_{2}$ solution, does the pH increase, decrease, or stay the same?

The Reasoning:

The Common Ion Effect: Adding $N a N O_{2}$ adds more $N O_{2}^{-}$ (the conjugate base).
Le Châtelier’s Principle: Increasing the concentration of a product ( $N O_{2}^{-}$ ) shifts the equilibrium to the left (toward the reactants).
The Result: This shift consumes $H^{+}$ ions. Since $[H^{+}]$ decreases, the pH increases.

Part 5: Selecting an Indicator

The Task: The equivalence point of this titration occurs at $p H = 8.1$ . Which indicator is best?

Methyl Red ( $p K_{a} = 5.1$ )
Bromthymol Blue ( $p K_{a} = 7.1$ )
Phenolphthalein ( $p K_{a} = 9.3$ )

The Solution:

Pick Phenolphthalein.
Justification: You want an indicator whose $p K_{a}$ is closest to the pH at the equivalence point. While 7.1 is numerically closer to 8.1, in a weak acid/strong base titration, the pH rises so sharply at the equivalence point that the range of Phenolphthalein ( $8.2 - 10$ ) perfectly captures the “end point” where the color change happens.

Final Master Tip for Tomorrow:

If you see a question about Percent Ionization, remember that as you dilute a weak acid (add water), the percent ionization increases, even though the total concentration of $H^{+}$ decreases. It’s a classic multiple-choice trap!

Would you like me to give you a “1-Minute Cheat Sheet” for the different types of Titration Curves you might see on the graph section?

🪴 Jacob's Notes

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