Acid-base reactions — neutralization, salt formation: Exercises

The single engine you will use everywhere:

Why does $M \cdot V$ equal moles? (build it before you use it)

Look at the picture below. Molarity $M$ means "moles packed into every one litre." So if you have $V$ litres, you simply have $M$ moles repeated $V$ times:

$mol per litre M \times litres V = moles (\frac{mol}{L} \times L = mol)$

The units literally cancel to leave "mol" — that unit cancellation is the proof the formula is right. This is the whole idea of Molarity and solution stoichiometry, drawn here as stacked boxes.

Figure — Acid-base reactions — neutralization, salt formation

Now multiply moles of acid by $n_{a}$ (protons each acid unit carries) to get moles of $H^{+}$ , and moles of base by $n_{b}$ to get moles of $O H^{-}$ . The next figure shows the endpoint as a balance: the two piles must be equal.

Read the balance-beam figure above. The left blue pan holds $n_{a} M_{a} V_{a}$ (all the protons); the right orange pan holds $n_{b} M_{b} V_{b}$ (all the hydroxides). When the green arrow says the beam is level — piles equal — every proton has found a partner hydroxide, and that is exactly the endpoint. Notice the beam does not care about volume alone or molarity alone; only the product with the $n$ -factor sits in each pan. This is why the diprotic-acid problems later "tip the beam" unless you remember the 2.

Why is $Δ H_{neut}$ a fixed $- 57.3 kJ/mol$ ? (build it too)

For any strong acid + strong base, strip out the spectator ions and the only bond that actually forms is the O–H bond of water: $H_{(a q)}^{+} + O H_{(a q)}^{-} \to H_{2} O_{(l)}$ Since this one identical reaction is the only chemical change every single time, the energy released per mole of water is always the same number. Change the acid or the base and you only change the spectators, which do nothing energetically — so the heat per mole cannot change. That constant is the fingerprint that proves only $H^{+} + O H^{-}$ reacts. See Enthalpy of reaction for the calorimetry.

Level 1 — Recognition

L1.1

Q: For the reaction $H N O_{3} + K O H \to ?$ , identify (a) the Brønsted acid, (b) the Brønsted base, (c) the salt formed, (d) the other product.

Recall Solution

(a) Acid = $H N O_{3}$ — it donates a proton ( $H^{+}$ ). See Brønsted-Lowry conjugate acid-base pairs.
(b) Base = $K O H$ — its $O H^{-}$ accepts the proton.
(c) Salt: cation from base ( $K^{+}$ ) + anion from acid ( $N O_{3}^{-}$ ) → $K N O_{3}$ (potassium nitrate). Charges are both $\pm 1$ , so a 1:1 pairing.
(d) The other product is water, $H_{2} O$ .
Full equation: $H N O_{3} + K O H \to K N O_{3} + H_{2} O$

L1.2

Q: State the net ionic equation for any strong acid reacting with any strong base, and say why it is the same for all of them.

Recall Solution

$H_{(a q)}^{+} + O H_{(a q)}^{-} \to H_{2} O_{(l)}$ Strong acids and bases are fully dissociated into free ions in water. The cation (e.g. $N a^{+}$ ) and the acid's anion (e.g. $C l^{-}$ ) are spectators — present before and after, unchanged. Strip them out and the only chemical change left is $H^{+}$ joining $O H^{-}$ to make water. Because that single reaction is identical every time, its enthalpy is a fixed $\approx - 57.3 kJ/mol$ (see Enthalpy of reaction).

Level 2 — Application

L2.1

Q: 20.0 mL of HCl is exactly neutralized by 25.0 mL of 0.150 M NaOH. Find the molarity of the HCl.

Recall Solution

Both are monoprotic/monobasic: $n_{a} = 1$ , $n_{b} = 1$ . $n_{a} M_{a} V_{a} = n_{b} M_{b} V_{b} \Rightarrow M_{a} (20.0) = (0.150) (25.0)$ $M_{a} = \frac{( 0.150 ) ( 25.0 )}{20.0} = \frac{3.75}{20.0} = 0.1875 M$ (Volumes stay in mL on both sides, so mL cancels mL — no conversion needed here.) Sanity check: the acid occupies the smaller volume (20 mL vs 25 mL) yet must deliver the same number of moles of $H^{+}$ as the base delivers of $O H^{-}$ . Fewer millilitres carrying the same moles means the acid must be more concentrated than the base. And indeed $0.1875 M > 0.150 M$ ✓.

L2.2

Q: How many mL of 0.20 M NaOH are needed to neutralize 30.0 mL of 0.10 M $H_{2} S O_{4}$ ?

Recall Solution

Sulfuric acid is diprotic: $n_{a} = 2$ . NaOH: $n_{b} = 1$ . $n_{a} M_{a} V_{a} = n_{b} M_{b} V_{b} \Rightarrow 2 (0.10) (30.0) = 1 (0.20) V_{b}$ $6.0 = 0.20 V_{b} \Rightarrow V_{b} = \frac{6.0}{0.20} = 30.0 mL$ Sanity: the acid supplies $2 \times 0.10 \times 30 = 6.0$ mmol of $H^{+}$ ; the base is twice as concentrated but supplies one $O H^{-}$ each, so equal 30 mL delivers $0.20 \times 30 = 6.0$ mmol $O H^{-}$ . Balanced ✓.

L2.3

Q: Write the balanced molecular equation for $H_{3} P O_{4} + N a O H$ (complete neutralization), and state $n_{a}$ .

Recall Solution

Phosphoric acid has 3 replaceable $H^{+}$ → $n_{a} = 3$ . Complete neutralization needs 3 $O H^{-}$ : $H_{3} P O_{4} + 3 N a O H \to N a_{3} P O_{4} + 3 H_{2} O$ Salt: cation $N a^{+}$ (charge $+ 1$ ) with $P O_{4}^{3 -}$ (charge $- 3$ ) → three $N a^{+}$ per phosphate → $N a_{3} P O_{4}$ , sodium phosphate.

Level 3 — Analysis

L3.1

Q: Predict whether an aqueous solution of $N H_{4} C l$ is acidic, basic, or neutral, and justify with an equation.

Recall Solution

$N H_{4} C l$ comes from the weak base $N H_{3}$ (as $N H_{4} O H$ ) + the strong acid $H C l$ . The cation $N H_{4}^{+}$ is the conjugate acid of a weak base, so it donates a proton to water: $N H_{4}^{+} + H_{2} O ⇌ N H_{3} + H_{3} O^{+}$ Extra $H_{3} O^{+}$ → the solution is acidic (pH < 7). This is Salt hydrolysis. The $C l^{-}$ (conjugate base of a strong acid) does not hydrolyze — it is a spectator.

L3.2

Q: A solution of $C H_{3} C O O N a$ turns out basic. Which parent was weak, and write the hydrolysis reaction.

Recall Solution

$C H_{3} C O O N a$ = weak acid $C H_{3} C O O H$ (acetic) + strong base $N a O H$ . The weak parent is the acid. Its conjugate base, acetate $C H_{3} C O O^{-}$ , grabs a proton from water: $C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$ Released $O H^{-}$ → basic (pH > 7). Rule: strong-base + weak-acid salt ⇒ basic. See pH and pOH scale.

L3.3

Q: 15.0 mL of 0.200 M $B a (O H)_{2}$ is titrated with 0.150 M HCl. What volume of HCl reaches the endpoint?

Recall Solution

$B a (O H)_{2}$ supplies 2 $O H^{-}$ → $n_{b} = 2$ . HCl: $n_{a} = 1$ . $n_{a} M_{a} V_{a} = n_{b} M_{b} V_{b} \Rightarrow 1 (0.150) V_{a} = 2 (0.200) (15.0)$ $0.150 V_{a} = 6.0 \Rightarrow V_{a} = \frac{6.0}{0.150} = 40.0 mL$ Reasoning: the base delivers $2 \times 0.200 \times 15 = 6.0$ mmol $O H^{-}$ , so we need 6.0 mmol $H^{+}$ ; from 0.150 M acid that is $6.0/0.150 = 40.0$ mL.

L3.4 (amphiprotic edge case)

Q: Sodium bicarbonate, $N a H C O_{3}$ , dissolves to give the $H C O_{3}^{-}$ ion. This ion can both donate a proton and accept one. Explain the two competing reactions and state whether the solution is (mildly) acidic or basic.

Recall Solution

An amphiprotic ion is one that can act as an acid or a base. For $H C O_{3}^{-}$ the two competing routes are:

As an acid (donate a proton): $H C O_{3}^{-} ⇌ H^{+} + C O_{3}^{2 -}$ .
As a base (accept a proton from water): $H C O_{3}^{-} + H_{2} O ⇌ H_{2} C O_{3} + O H^{-}$ . Which wins? Compare the tendencies: $H C O_{3}^{-}$ is a very weak acid (it holds its proton tightly) but a relatively stronger base (it grabs protons more readily). The base route dominates, so slightly more $O H^{-}$ is produced → the solution is mildly basic (pH ≈ 8). The same logic applies to $H P O_{4}^{2 -}$ . This is an extension of Salt hydrolysis — the difference is that here the same ion plays both roles.

Level 4 — Synthesis

L4.1

Q: A 0.500 g sample of impure NaOH is dissolved and requires 22.5 mL of 0.200 M HCl for complete neutralization. What is the percent purity of the NaOH? (Molar mass NaOH = 40.0 g/mol.)

Recall Solution

First convert the volume to litres so it pairs with molarity (mol/L): $22.5 mL = 22.5/1000 = 0.0225 L$ .
Moles of $H^{+}$ from acid = $M V = 0.200 \times 0.0225 = 0.00450$ mol. Since $n_{a} = n_{b} = 1$ , this equals moles of NaOH that reacted.
Mass of pure NaOH = $0.00450 \times 40.0 = 0.180$ g.
Percent purity $= \frac{0.180}{0.500} \times 100 = 36.0%$ . Interpretation: only 36% of the sample was actual base; the rest was inert impurity.

L4.2

Q: 50.0 mL of 0.100 M $H_{2} S O_{4}$ is mixed with 50.0 mL of 0.100 M NaOH. Is the final mixture acidic, basic, or neutral? Find the moles of excess ion.

Recall Solution

Convert volumes: $50.0 mL = 0.0500 L$ (both solutions).
$H^{+}$ available: $n_{a} M_{a} V_{a} = 2 (0.100) (0.0500) = 0.0100$ mol.
$O H^{-}$ available: $n_{b} M_{b} V_{b} = 1 (0.100) (0.0500) = 0.00500$ mol.
They react 1:1, so $0.00500$ mol $H^{+}$ is consumed, leaving excess $H^{+} = 0.0100 - 0.00500 = 0.00500 mol$ .
Excess $H^{+}$ ⇒ the mixture is acidic. Lesson: equal volume + equal molarity did not neutralize, because the diprotic acid brought twice the protons.

L4.3

Q: What mass of $K O H$ (molar mass 56.1 g/mol) is needed to fully neutralize 40.0 mL of 0.250 M $H_{3} P O_{4}$ ?

Recall Solution

Convert the volume: $40.0 mL = 0.0400 L$ .
$H^{+}$ from acid: $n_{a} M_{a} V_{a} = 3 (0.250) (0.0400) = 0.0300$ mol.
$O H^{-}$ needed = $0.0300$ mol; since $n_{b} = 1$ , moles of $K O H = 0.0300$ mol.
Mass $= 0.0300 \times 56.1 = 1.683 g$ .

Level 5 — Mastery

L5.1 (double titration / back-titration flavor)

Q: 25.0 mL of a solution containing both HCl and $H_{2} S O_{4}$ requires 30.0 mL of 0.200 M NaOH for complete neutralization. If the HCl contributes 0.00200 mol of $H^{+}$ , find the molarity of $H_{2} S O_{4}$ in the original solution.

Recall Solution

Convert volumes: NaOH $= 30.0 mL = 0.0300 L$ ; sample $= 25.0 mL = 0.0250 L$ .
Total $O H^{-}$ supplied = $M V = 0.200 \times 0.0300 = 0.00600$ mol. This equals total $H^{+}$ from both acids.
$H^{+}$ from HCl = $0.00200$ mol, so $H^{+}$ from $H_{2} S O_{4} = 0.00600 - 0.00200 = 0.00400$ mol.
Each $H_{2} S O_{4}$ gives 2 $H^{+}$ , so moles of $H_{2} S O_{4} = 0.00400/2 = 0.00200$ mol.
Molarity $= \frac{0.00200}{0.0250} = 0.0800 M$ .

L5.2 (enthalpy tie-in)

Q: When 100.0 mL of 0.500 M HCl is mixed with 100.0 mL of 0.500 M NaOH, estimate the heat released, using $Δ H = - 57.3 kJ/mol$ for $H^{+} + O H^{-} \to H_{2} O$ .

Recall Solution

Convert volumes: each $= 100.0 mL = 0.1000 L$ .
Moles of $H^{+}$ = $0.500 \times 0.1000 = 0.0500$ mol; moles of $O H^{-}$ = same = $0.0500$ mol → exact match, all react.
Moles of water formed = $0.0500$ mol.
Heat released $= 0.0500 \times 57.3 = 2.865 kJ$ (i.e. $Δ H = - 2.865 kJ$ ). Why the fixed 57.3: only $H^{+} + O H^{-}$ reacts, so heat scales purely with moles of water — see Enthalpy of reaction.

L5.3 (indicator / equivalence-point reasoning)

Q: In titrating acetic acid ( $C H_{3} C O O H$ , weak) with NaOH (strong), is the pH at the equivalence point equal to, above, or below 7? Which broad indicator class (acidic-range vs basic-range) suits it?

Recall Solution

At the equivalence point all acetic acid has become sodium acetate, $C H_{3} C O O N a$ . As in L3.2, acetate hydrolyzes: $C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}$ .
Extra $O H^{-}$ ⇒ equivalence-point pH is above 7 (basic).
You need an indicator that changes color in the basic range (e.g. phenolphthalein, ~8–10), not one that flips in acidic range. See Titration and indicators. Key idea: the equivalence point (moles matched) and the neutral point (pH 7) are not the same when a weak partner is involved.

Wrap-up recall

Recall One-line takeaways (hide first)

Master formula ::: $n_{a} M_{a} V_{a} = n_{b} M_{b} V_{b}$
Why $H_{2} S O_{4}$ needs double NaOH ::: it is diprotic, $n_{a} = 2$
Salt of strong base + weak acid ::: basic (anion hydrolyzes)
Salt of strong acid + weak base ::: acidic (cation hydrolyzes)
An amphiprotic ion like $H C O_{3}^{-}$ is ::: mildly basic (its base tendency beats its acid tendency)
Heat of strong–strong neutralization scales with ::: moles of water formed ( $\times 57.3$ kJ/mol)

Connections

Parent topic
Molarity and solution stoichiometry
Titration and indicators
Salt hydrolysis
pH and pOH scale
Brønsted-Lowry conjugate acid-base pairs
Enthalpy of reaction