What is Titration on the Chemistry Regents?

Titration is a Solutions & Acids/Bases topic on the Chemistry Regents. RegentsLab has 32 practice questions on it, each with a full worked solution.

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Titration: Chemistry Regents Practice

Practice Titration for the Chemistry Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 7 · 2 pts

A $20.0\ \text{mL}$ sample of $\text{HCl}(aq)$ is exactly neutralized by $25.0\ \text{mL}$ of $0.100\ \text{M}$ $\text{NaOH}(aq)$. What is the molarity of the $\text{HCl}(aq)$?

(A) $0.0800\ \text{M}$
(B) $0.125\ \text{M}$ ✓
(C) $0.100\ \text{M}$
(D) $2.50\ \text{M}$

Worked solution

$M_A V_A = M_B V_B$ Table T titration relationship.

$M_A = \frac{(0.100)(25.0)}{20.0}$ Solve for the unknown acid molarity.

$M_A = 0.125\ \text{M}$

Answer: 0.125\ \text{M}

At neutralization the moles of $\text{H}^+$ equal the moles of $\text{OH}^-$, so rearranging Table T's $M_AV_A = M_BV_B$ gives the unknown molarity.

Question 8 · 2 pts

What volume of $0.50\ \text{M}$ $\text{NaOH}(aq)$ is required to exactly neutralize $30.0\ \text{mL}$ of $0.25\ \text{M}$ $\text{HCl}(aq)$?

(A) $60.\ \text{mL}$
(B) $30.\ \text{mL}$
(C) $3.75\ \text{mL}$
(D) $15\ \text{mL}$ ✓

Worked solution

$M_A V_A = M_B V_B$ Table T; solve for the base volume.

$V_B = \frac{(0.25)(30.0)}{0.50}$

$V_B = 15\ \text{mL}$

Answer: 15\ \text{mL}

Since the base is twice as concentrated as the acid, it takes half the volume of base to deliver the same number of moles of $\text{OH}^-$.

Question 56 · 1 pt

$40.0\ \mathrm{mL}$ of $\mathrm{HCl}$ is neutralized by $20.0\ \mathrm{mL}$ of $0.20\ \mathrm{M}\ \mathrm{NaOH}$. What is the molarity of the acid?

(A) $0.40\ \mathrm{M}$
(B) $4.0\ \mathrm{M}$
(C) $0.10\ \mathrm{M}$ ✓
(D) $0.20\ \mathrm{M}$

Worked solution

$M_a V_a = M_b V_b$

$M_a(40.0) = (0.20)(20.0)$

$M_a = \frac{4.0}{40.0} = 0.10\ \mathrm{M}$

Answer: 0.10\ \mathrm{M}

Set moles of acid equal to moles of base, then solve for the unknown molarity.

Question 9 · 2 pts

During an acid-base titration, the endpoint is the moment best identified by

(A) a color change of the indicator ✓
(B) the solution boiling
(C) the formation of a precipitate
(D) a gas bubbling out of the flask

Worked solution

$\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ Neutralization forms a soluble salt and water.

$\text{added moles OH}^- = \text{moles H}^+ \Rightarrow \text{indicator changes color}$ An indicator is chosen to change color right at this point.

Answer: a color change of the indicator

A titration adds a known solution until an indicator changes color, signaling the endpoint where the acid has been neutralized by the base.

Question 10 · 2 pts

A $15.0\ \text{mL}$ sample of $0.200\ \text{M}$ $\text{HCl}(aq)$ is exactly neutralized by $25.0\ \text{mL}$ of $\text{KOH}(aq)$. What is the molarity of the $\text{KOH}(aq)$?

(A) $0.200\ \text{M}$
(B) $0.333\ \text{M}$
(C) $3.00\ \text{M}$
(D) $0.120\ \text{M}$ ✓

Worked solution

$M_A V_A = M_B V_B$ Table T; solve for the base molarity.

$M_B = \frac{(0.200)(15.0)}{25.0}$

$M_B = 0.120\ \text{M}$

Answer: 0.120\ \text{M}

Rearranging $M_AV_A = M_BV_B$ isolates the base molarity; the larger base volume means a molarity smaller than the acid's.

Question 12 · 2 pts

In a titration, 25.0 mL of 0.20 M HCl exactly neutralizes 40.0 mL of a NaOH solution. What is the molarity of the NaOH solution?

(A) 0.125 M ✓
(B) 0.32 M
(C) 0.80 M
(D) 0.031 M

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ 1:1 mole ratio for HCl and NaOH

$(0.20)(25.0) = M_{base}(40.0)$

$M_{base} = \dfrac{5.0}{40.0} = 0.125\ M$

Answer: 0.125 M

Because HCl and NaOH react in a 1:1 mole ratio, the moles of acid at neutralization equal the moles of base, so M_acid times V_acid equals M_base times V_base. Solving for M_base gives 0.125 M. Swapping the volumes in the equation produces the 0.32 M distractor.

Question 31 · 2 pts

In a titration, 25.0 mL of 0.20 M $HCl$ exactly neutralizes 50.0 mL of $NaOH$ solution. What is the molarity of the $NaOH$ solution?

(A) 0.20 M
(B) 0.40 M
(C) 0.05 M
(D) 0.10 M ✓

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ 1:1 mole ratio for HCl and NaOH

$(0.20\ M)(25.0\ mL) = M_{base}(50.0\ mL)$

$M_{base} = \dfrac{5.0}{50.0} = 0.10\ M$

Answer: 0.10 M

Since HCl and NaOH react in a 1:1 mole ratio, $M_1V_1 = M_2V_2$ applies; solving (0.20)(25.0) = M(50.0) gives M = 0.10 M for the NaOH solution.

Question 70 · 4 pts · constructed response

In a titration, 25.0 mL of an HCl solution is exactly neutralized by 40.0 mL of 0.50 M NaOH solution. The neutralization reaction is $HCl + NaOH \rightarrow NaCl + H_2O$. Determine the molarity of the HCl solution.

Model answer: The molarity of the HCl solution is 0.80 M.

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ Titration formula for a 1:1 acid-base reaction

$M_{acid}(25.0\ mL) = (0.50\ M)(40.0\ mL)$ Substitute given values

$M_{acid} = \dfrac{(0.50)(40.0)}{25.0} = 0.80\ M$ Solve for the acid molarity

Answer: The molarity of the HCl solution is 0.80 M.

Because HCl and NaOH react in a 1:1 mole ratio, the moles of acid at the equivalence point equal the moles of base, so MaVa = MbVb can be used directly to solve for the unknown molarity.

Question 85 · 4 pts · constructed response

A student titrates $25.0\ mL$ of an HCl solution of unknown concentration with $0.100\ M$ NaOH solution. It takes $30.0\ mL$ of the NaOH solution to reach the endpoint of the titration. The neutralization reaction is $HCl + NaOH \rightarrow NaCl + H_2O$. Calculate the molarity of the HCl solution.

Model answer: The molarity of the HCl solution is $0.120\ M$.

Worked solution

$M_{HCl}V_{HCl} = M_{NaOH}V_{NaOH}$ 1:1 mole ratio of acid to base in the balanced equation

$M_{HCl}(25.0\ mL) = (0.100\ M)(30.0\ mL)$

$M_{HCl} = \dfrac{(0.100)(30.0)}{25.0} = 0.120\ M$

Answer: The molarity of the HCl solution is $0.120\ M$.

At the endpoint of a titration, moles of acid equal moles of base when the mole ratio is 1:1, so $M_{acid}V_{acid} = M_{base}V_{base}$ can be used to find the unknown concentration.

Question 85 · 4 pts · constructed response

A 25.0-milliliter sample of $HCl(aq)$ is completely neutralized by 32.0 milliliters of 0.150 M $NaOH(aq)$, according to the equation $HCl(aq) + NaOH(aq) \rightarrow NaCl(aq) + H_2O(\ell)$. Determine the molarity of the $HCl(aq)$ solution.

Model answer: The molarity of the HCl solution is 0.192 M.

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ Valid because the mole ratio of acid to base in the equation is 1:1

$M_{acid}(25.0\ mL) = (0.150\ M)(32.0\ mL)$

$M_{acid} = \dfrac{(0.150)(32.0)}{25.0} = 0.192\ M$

Answer: The molarity of the HCl solution is 0.192 M.

Since HCl and NaOH react in a 1:1 mole ratio, the titration relationship $M_{acid}V_{acid} = M_{base}V_{base}$ can be used directly to solve for the unknown molarity using the volumes and known molarity given.

Question 24 · 2 pts

In a titration, 25.0 mL of 0.20 M NaOH exactly neutralizes 50.0 mL of an HCl solution. What is the molarity of the HCl solution?

(A) 0.20 M
(B) 0.10 M ✓
(C) 0.050 M
(D) 0.40 M

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ Titration formula, both are monoprotic

$M_{acid} \times 50.0\ \text{mL} = 0.20\ \text{M} \times 25.0\ \text{mL}$

$M_{acid} = \dfrac{0.20 \times 25.0}{50.0} = 0.10\ \text{M}$

Answer: 0.10 M

For a titration of a monoprotic acid with a monoprotic base, moles of acid equal moles of base at the equivalence point, giving $M_aV_a = M_bV_b$. Solving for the acid molarity with the given volumes yields 0.10 M HCl.

Question 59 · 4 pts · constructed response

In a titration, 25.0 mL of 0.150 M HCl is exactly neutralized by 30.0 mL of NaOH solution of unknown concentration, according to the equation $HCl + NaOH \rightarrow NaCl + H_2O$. Calculate the molarity of the NaOH solution.

Model answer: 0.125 M

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ Since the acid-base mole ratio is 1:1, moles of HCl equal moles of NaOH at neutralization

$(0.150\ \text{M})(25.0\ \text{mL}) = M_{base}(30.0\ \text{mL})$

$M_{base} = \dfrac{(0.150)(25.0)}{30.0} = 0.125\ \text{M}$

Answer: 0.125 M

At the equivalence point of a titration, moles of acid equal moles of base for a 1:1 reaction, so setting M_aV_a equal to M_bV_b and solving for the unknown molarity gives the NaOH concentration.

Question 58 · 1 pt

During a titration, $25.0\ \mathrm{mL}$ of $\mathrm{HCl}$ is exactly neutralized by $30.0\ \mathrm{mL}$ of $0.10\ \mathrm{M}\ \mathrm{NaOH}$. What is the molarity of the $\mathrm{HCl}$ ?

(A) $0.10\ \mathrm{M}$
(B) $3.0\ \mathrm{M}$
(C) $0.083\ \mathrm{M}$
(D) $0.12\ \mathrm{M}$ ✓

Worked solution

$M_a V_a = M_b V_b$ 1:1 titration (Reference Tables).

$M_a(25.0) = (0.10)(30.0)$

$M_a = \frac{3.0}{25.0} = 0.12\ \mathrm{M}$

Answer: 0.12\ \mathrm{M}

At the equivalence point of a 1:1 acid–base titration, moles of acid equal moles of base: MₐVₐ = M_bV_b. Solve for the unknown and divide by its matching volume.

Question 35 · 1 pt

In a titration, $20.0$ mL of $HCl$ is neutralized by $40.0$ mL of $0.50\ M\ NaOH$. What is the molarity of the acid?

(A) $2.0\ M$
(B) $0.25\ M$
(C) $1.0\ M$ ✓
(D) $0.50\ M$

Worked solution

$M_aV_a=M_bV_b$ 1:1 acid–base, from the Reference Tables.

$M_a=\frac{(0.50)(40.0)}{20.0}=1.0\ M$

Answer: 1.0\ M

For a 1:1 neutralization, MₐVₐ = M_bV_b; solve for the acid's molarity.

Question 11 · 2 pts

In a titration, $50.0\ \text{mL}$ of $\text{HCl}(aq)$ is exactly neutralized by $40.0\ \text{mL}$ of $0.25\ \text{M}$ $\text{NaOH}(aq)$. What is the total number of moles of $\text{HCl}$ neutralized?

(A) $10.\ \text{mol}$
(B) $0.20\ \text{mol}$
(C) $0.010\ \text{mol}$ ✓
(D) $0.0125\ \text{mol}$

Worked solution

$n_{\text{NaOH}} = M \times V = 0.25 \times 0.0400$ Convert 40.0 mL to 0.0400 L first.

$n_{\text{NaOH}} = 0.010\ \text{mol}$

$n_{\text{HCl}} = n_{\text{NaOH}} = 0.010\ \text{mol}$ At neutralization moles H+ = moles OH-.

Answer: 0.010\ \text{mol}

Moles equal molarity times volume in liters, and at the equivalence point the moles of acid equal the moles of base added.

Question 12 · 2 pts

Two $20.0\ \text{mL}$ acid samples are each titrated with the same $0.100\ \text{M}$ $\text{NaOH}(aq)$. Sample X requires $20.0\ \text{mL}$ of the base; Sample Y requires $40.0\ \text{mL}$. Which statement is correct?

(A) Sample Y is four times as concentrated as Sample X.
(B) The two samples have equal concentration.
(C) Sample X is twice as concentrated as Sample Y.
(D) Sample Y is twice as concentrated as Sample X. ✓

Worked solution

$M_X = \frac{(0.100)(20.0)}{20.0} = 0.100\ \text{M}$ Table T for Sample X.

$M_Y = \frac{(0.100)(40.0)}{20.0} = 0.200\ \text{M}$ Table T for Sample Y.

$\frac{M_Y}{M_X} = \frac{0.200}{0.100} = 2$ Y is twice as concentrated.

Answer: Sample Y is twice as concentrated as Sample X.

With equal acid volumes, the acid needing more titrant is more concentrated; doubling the base volume required doubles the calculated acid molarity.

Question 43 · 2 pts

In a titration, 25.0 mL of 0.10 M HCl exactly neutralizes 50.0 mL of a NaOH solution of unknown concentration. What is the molarity of the NaOH solution?

(A) 0.025 M
(B) 0.10 M
(C) 0.20 M
(D) 0.050 M ✓

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ At the equivalence point, moles of H+ equal moles of OH- for a 1:1 acid-base reaction

$(0.10\ M)(25.0\ mL) = M_{base}(50.0\ mL)$

$M_{base} = \dfrac{(0.10)(25.0)}{50.0} = 0.050\ M$

Answer: 0.050 M

Since HCl and NaOH react in a 1:1 mole ratio, the titration formula M_acid V_acid = M_base V_base applies, giving a NaOH molarity of 0.050 M. A common error is dividing volumes in the wrong order or forgetting to account for the larger base volume, which dilutes the concentration relative to the acid.

Question 58 · 4 pts · constructed response

In a titration, 25.0 mL of an HCl solution of unknown molarity is exactly neutralized by 30.0 mL of 0.150 M NaOH solution, according to the equation $HCl + NaOH \rightarrow NaCl + H_2O$. Calculate the molarity of the HCl solution.

Model answer: The molarity of the HCl solution is 0.180 M.

Worked solution

$M_{a}V_{a} = M_{b}V_{b}$ Valid because the acid and base react in a 1:1 mole ratio

$M_{a}(25.0\ \text{mL}) = (0.150\ \text{M})(30.0\ \text{mL})$

$M_{a} = \dfrac{(0.150)(30.0)}{25.0} = 0.180\ \text{M}$

Answer: The molarity of the HCl solution is 0.180 M.

At the equivalence point of a titration, moles of acid equal moles of base when the mole ratio of acid to base is 1:1, allowing the unknown concentration to be found using the relationship M_aV_a = M_bV_b.

Question 66 · 4 pts · constructed response

In a titration, 25.0 mL of an HCl solution of unknown concentration is exactly neutralized by 30.0 mL of 0.20 M NaOH solution, according to the equation $HCl + NaOH \rightarrow NaCl + H_2O$. Determine the molarity of the HCl solution.

Model answer: 0.24 M

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ 1:1 mole ratio between HCl and NaOH

$M_{acid} = \dfrac{M_{base}V_{base}}{V_{acid}} = \dfrac{(0.20)(30.0)}{25.0}$

$M_{acid} = 0.24 \text{ M}$

Answer: 0.24 M

Because HCl and NaOH react in a 1:1 mole ratio, the moles of acid neutralized equal the moles of base added, allowing the titration formula MaVa = MbVb to be used to solve for the unknown acid molarity.

Question 82 · 4 pts · constructed response

In a titration, a 25.0 mL sample of HCl solution of unknown concentration is exactly neutralized by 30.0 mL of a 0.20 M NaOH solution. Given that HCl and NaOH react in a 1:1 mole ratio, determine the molarity of the HCl solution.

Model answer: The molarity of the HCl solution is 0.24 M.

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ Valid since HCl and NaOH react 1:1

$M_{acid}(25.0 \text{ mL}) = (0.20 \text{ M})(30.0 \text{ mL})$

$M_{acid} = \dfrac{(0.20)(30.0)}{25.0} = 0.24 \text{ M}$

Answer: The molarity of the HCl solution is 0.24 M.

At the equivalence point of a titration, the moles of acid equal the moles of base when the reaction ratio is 1:1, so M_acid × V_acid = M_base × V_base. Solving for the unknown molarity using the given volumes and known base concentration gives 0.24 M.

Question 10 · 2 pts

In a titration, 25.0 mL of 0.10 M HCl exactly neutralizes 20.0 mL of an NaOH solution of unknown concentration. What is the molarity of the NaOH solution?

(A) 0.20 M
(B) 0.125 M ✓
(C) 0.080 M
(D) 0.10 M

Worked solution

$HCl + NaOH \rightarrow NaCl + H_2O$ 1:1 mole ratio of acid to base

$M_aV_a = M_bV_b$

$(0.10)(25.0) = M_b(20.0)$

$M_b = \dfrac{2.5}{20.0} = 0.125\ M$

Answer: 0.125 M

Since HCl and NaOH react in a 1:1 mole ratio, the titration formula MaVa = MbVb gives the NaOH molarity as (0.10 × 25.0) ÷ 20.0 = 0.125 M.

Question 24 · 2 pts

In a titration, 25.0 mL of 0.20 M HCl exactly neutralizes 50.0 mL of a NaOH solution. What is the molarity of the NaOH solution?

(A) 0.40 M
(B) 0.050 M
(C) 0.10 M ✓
(D) 0.20 M

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ HCl and NaOH react in a 1:1 mole ratio

$(0.20\ \text{M})(25.0\ \text{mL}) = M_{base}(50.0\ \text{mL})$

$M_{base} = \frac{5.0}{50.0} = 0.10\ \text{M}$

Answer: 0.10 M

Because HCl and NaOH neutralize in a 1:1 mole ratio, the moles of acid equal the moles of base at the equivalence point, so MaVa = MbVb. Solving for Mb using the given volumes and acid molarity gives 0.10 M. Forgetting to account for the larger base volume is a common source of error.

Question 34 · 2 pts

In a titration, 25.0 mL of an HCl solution of unknown concentration is exactly neutralized by 30.0 mL of 0.20 M NaOH, according to the equation $HCl + NaOH \rightarrow NaCl + H_2O$. What is the concentration of the HCl solution?

(A) 0.30 M
(B) 0.17 M
(C) 0.20 M
(D) 0.24 M ✓

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ 1:1 mole ratio of HCl to NaOH in the balanced equation

$M_{acid}(25.0\text{ mL}) = (0.20\text{ M})(30.0\text{ mL})$

$M_{acid} = \dfrac{(0.20)(30.0)}{25.0} = 0.24\text{ M}$

Answer: 0.24 M

Because HCl and NaOH react in a 1:1 mole ratio, the moles of acid at the equivalence point equal the moles of base. Setting M-acid times V-acid equal to M-base times V-base and solving gives an HCl concentration of 0.24 M.

Question 9 · 2 pts

In a titration, 25.0 mL of an HCl solution of unknown concentration is exactly neutralized by 20.0 mL of 0.100 M NaOH solution. What is the concentration of the HCl solution? ($HCl + NaOH \rightarrow NaCl + H_2O$)

(A) 0.080 M ✓
(B) 0.100 M
(C) 0.500 M
(D) 0.125 M

Worked solution

$M_{acid}V_{acid} = M_{base}V_{base}$ 1:1 mole ratio from the balanced equation

$M_{acid}(25.0 \text{ mL}) = (0.100 \text{ M})(20.0 \text{ mL})$

$M_{acid} = \frac{(0.100)(20.0)}{25.0} = 0.080 \text{ M}$

Answer: 0.080 M

Since HCl and NaOH react in a 1:1 mole ratio, the moles of acid equal the moles of base at the equivalence point, so M(acid)V(acid) = M(base)V(base). Solving for the unknown molarity gives 0.080 M.

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