In an acid-base back-titration experiment, a student dissolves an impure $1.500\text{ g}$ sample containing magnesium hydroxide into $50.00\text{ mL}$ of $0.5000\text{ M } \text{HCl}$. The unreacted excess acid requires $18.40\text{ mL}$ of $0.2500\text{ M } \text{NaOH}$ to reach the phenolphthalein equivalence point according to the neutralization reactions $\text{Mg(OH)}_2 + 2\text{HCl} \to \text{MgCl}_2 + 2\text{H}_2\text{O}$ and $\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}$. Set up and solve the multi-step stoichiometric equation to calculate the mass (in grams) of pure $\text{Mg(OH)}_2$ (molar mass $58.32\text{ g/mol}$) present in the sample.

In an acid-base back-titration experiment, a student dissolves an impure $1.500\text{ g}$ sample containing magnesium hydroxide into $50.00\text{ mL}$ of $0.5000\text{ M } \text{HCl}$ . The unreacted excess acid requires $18.40\text{ mL}$ of $0.2500\text{ M } \text{NaOH}$ to reach the phenolphthalein equivalence point according to the neutralization reactions $\text{Mg(OH)}_2 + 2\text{HCl} \to \text{MgCl}_2 + 2\text{H}_2\text{O}$ and $\text{HCl} + \text{NaOH} \to \text{NaCl} + \text{H}_2\text{O}$ . Set up and solve the multi-step stoichiometric equation to calculate the mass (in grams) of pure $\text{Mg(OH)}_2$ (molar mass $58.32\text{ g/mol}$ ) present in the sample.

Final answer

$0.5949\text{ g}$

Steps

Calculate the total moles of $\ce{HCl}$ added

To find the mass of pure $\ce{Mg(OH)2}$ in the sample, we use the principle of back-titration: the total moles of acid added equal the moles of acid consumed by the base sample plus the moles of acid neutralized by the titrant ( $\ce{NaOH}$ ).

Using the formula $n = C \cdot V$ : $n(\ce{HCl})_{\text{total}} = 0.5000\text{ M} \times 0.05000\text{ L} = 0.02500\text{ mol}$

Calculate the unreacted (excess) moles of $\ce{HCl}$

The excess $\ce{HCl}$ is titrated to the equivalence point with $\ce{NaOH}$ : $\ce{HCl + NaOH -> NaCl + H2O}$

From the $1:1$ stoichiometric ratio: $n(\ce{HCl})_{\text{excess}} = n(\ce{NaOH}) = 0.2500\text{ M} \times 0.01840\text{ L} = 0.004600\text{ mol}$

Determine the moles of $\ce{HCl}$ that reacted with $\ce{Mg(OH)2}$

Subtract the unreacted acid from the total initial acid: $n(\ce{HCl})_{\text{reacted}} = n(\ce{HCl})_{\text{total}} - n(\ce{HCl})_{\text{excess}}$ $n(\ce{HCl})_{\text{reacted}} = 0.02500\text{ mol} - 0.004600\text{ mol} = 0.02040\text{ mol}$

Calculate the moles and mass of pure $\ce{Mg(OH)2}$

From the balanced equation: $\ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}$

Each mole of $\ce{Mg(OH)2}$ consumes $2\text{ mol}$ of $\ce{HCl}$ . Therefore: $n(\ce{Mg(OH)2}) = \frac{n(\ce{HCl})_{\text{reacted}}}{2} = \frac{0.02040\text{ mol}}{2} = 0.01020\text{ mol}$

Now, multiply by the molar mass of $\ce{Mg(OH)2}$ ( $58.32\text{ g/mol}$ ): $\text{Mass of } \ce{Mg(OH)2} = 0.01020\text{ mol} \times 58.32\text{ g/mol} = 0.594864\text{ g}$

Rounding to four significant figures yields: $\text{Mass of } \ce{Mg(OH)2} = 0.5949\text{ g}$

Common Pitfall: Forgetting the $1:2$ stoichiometric ratio between $\ce{Mg(OH)2}$ and $\ce{HCl}$ is a frequent error. Always verify the balanced chemical equation before converting moles of acid consumed to moles of metal hydroxide.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student ignores the 1:2 stoichiometric ratio between Mg(OH)2 and HCl, assuming a 1:1 ratio instead.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$n(\text{total } \ce{HCl}) = 0.05000 \times 0.5000 = 0.02500\text{ mol}$
Step 2
$n(\text{excess } \ce{HCl}) = n(\ce{NaOH}) = 0.01840 \times 0.2500 = 0.004600\text{ mol}$
Step 3
$n(\text{reacted } \ce{HCl}) = 0.02500 - 0.004600 = 0.02040\text{ mol}$
Step 4
$n(\ce{Mg(OH)2}) = n(\text{reacted } \ce{HCl}) = 0.02040\text{ mol}$
Correct
$n(\ce{Mg(OH)2}) = \frac{1}{2} n(\ce{HCl})_{\text{reacted}} = \frac{0.02040\text{ mol}}{2} = 0.01020\text{ mol}$
From the balanced equation $\ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}$ , each mole of $\ce{Mg(OH)2}$ provides two hydroxide ions and neutralizes two moles of $\ce{HCl}$ . Therefore, $n(\ce{Mg(OH)2})$ must equal half of the reacted moles of acid.
Step 5
$m(\ce{Mg(OH)2}) = 0.02040 \times 58.32 = 1.190\text{ g}$
Correct
$m(\ce{Mg(OH)2}) = 0.01020\text{ mol} \times 58.32\text{ g/mol} = 0.5949\text{ g}$
You multiplied your mole value by the correct molar mass, but the resulting mass is doubled due to the stoichiometric error in Step 4.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly calculated the total initial moles of $\ce{HCl}$ .

You correctly calculated the moles of unreacted excess $\ce{HCl}$ from the sodium hydroxide titration.

You correctly determined the net moles of acid consumed by subtracting excess acid from total acid.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Stoichiometric Ratio Omission

You equated the moles of magnesium hydroxide directly to the moles of hydrochloric acid as if the reaction had a 1:1 mole ratio.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You overlooked the stoichiometric coefficient 2 in front of $\ce{HCl}$ in the neutralization equation $\ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}$ , treating the dihydroxide base as monoprotic.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check Hydroxide-to-Proton Ratio: For a polyprotic base $\ce{M(OH)}_n$ reacting with monoprotic acid $\ce{HX}$ , $n(\ce{HX}) = n \cdot n(\ce{M(OH)}_n)$ .
Always write down the balanced neutralization equation and convert moles of acid consumed using $\frac{n(\ce{acid})}{\text{stoichiometric coefficient}}$ .

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the balanced reaction stoichiometry: $n(\ce{Mg(OH)2}) = \frac{n(\ce{HCl})_{\text{reacted}}}{2} = \frac{0.02040\text{ mol}}{2} = 0.01020\text{ mol}$ .
Multiply the correct moles by the molar mass: $m = 0.01020\text{ mol} \times 58.32\text{ g/mol} = 0.594864\text{ g} \approx 0.5949\text{ g}$ .

Final answer

$0.5949\text{ g}$

NEXT STEPS

Final Takeaways from Your Teacher

Review balanced chemical equations specifically for metal hydroxides with multiple $\ce{OH-}$ groups.
Before multiplying by molar mass, explicitly check whether the mole ratio between the analyte and the reactant is $1:1$ , $1:2$ , or $1:3$ .

OverviewACEpath Teacher's Diagnosis

02 The student multiplies by 2 instead of dividing by 2 when converting moles of HCl to moles of Mg(OH)2.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$n(\text{total HCl}) = 0.05000 \times 0.5000 = 0.02500\text{ mol}$
Step 2
$n(\text{excess HCl}) = 0.01840 \times 0.2500 = 0.004600\text{ mol}$
Step 3
$n(\text{reacted HCl}) = 0.02500 - 0.004600 = 0.02040\text{ mol}$
Step 4
$n(\ce{Mg(OH)2}) = 0.02040 \times 2 = 0.04080\text{ mol}$
Correct
$n(\ce{Mg(OH)2}) = \frac{n(\text{reacted HCl})}{2} = \frac{0.02040\text{ mol}}{2} = 0.01020\text{ mol}$
From the balanced equation $\ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}$ , each mole of $\ce{Mg(OH)2}$ reacts with $2\text{ mol}$ of $\ce{HCl}$ . Therefore, the moles of $\ce{Mg(OH)2}$ equal $\frac{1}{2}$ the moles of reacted $\ce{HCl}$ , so you must divide by $2$ rather than multiply by $2$ .
Step 5
$m(\ce{Mg(OH)2}) = 0.04080 \times 58.32 = 2.379\text{ g}$
Correct
$m(\ce{Mg(OH)2}) = 0.01020\text{ mol} \times 58.32\text{ g/mol} = 0.5949\text{ g}$
Your molar mass calculation followed correctly from your previous value, but inherited the inverted stoichiometric factor. Notice also that $2.379\text{ g}$ exceeds the initial $1.500\text{ g}$ sample mass, which provides a helpful sanity check that the moles of analyte were overestimated.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly converted all titration volumes from milliliters to liters.

You accurately determined the total moles of $\ce{HCl}$ and the excess moles neutralized by $\ce{NaOH}$ .

You used the correct conservation relationship to find the net moles of $\ce{HCl}$ consumed by the sample.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Inverted Stoichiometric Ratio

Multiplied by the coefficient $2$ instead of dividing by $2$ when relating $\ce{HCl}$ to $\ce{Mg(OH)2}$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the direction of the stoichiometric conversion factor between the known substance ( $\ce{HCl}$ ) and the unknown target substance ( $\ce{Mg(OH)2}$ ).

REASONING PATH

Your ACEpath teacher’s suggested approach

Dimensional Analysis Check: Always write full units and species in the ratio: $\text{moles unknown} = \text{moles known} \times \frac{\text{coefficient of unknown}}{\text{coefficient of known}}$ .
Multiply $0.02040\text{ mol }\ce{HCl}$ by $\frac{1\text{ mol }\ce{Mg(OH)2}}{2\text{ mol }\ce{HCl}}$ to ensure the unit $\text{mol }\ce{HCl}$ cancels out.
Physical Reality Check: The mass of pure constituent extracted from an impure mixture cannot exceed the total mixture sample mass.
Compare your calculated mass $2.379\text{ g}$ to the sample mass $1.500\text{ g}$ ; exceeding $100\%$ purity signals an arithmetic or ratio error.

CORRECTED PATH

ACEpath Teacher's Demonstration

Apply the stoichiometric ratio from the balanced equation: $n(\ce{Mg(OH)2}) = \frac{0.02040\text{ mol}}{2} = 0.01020\text{ mol}$ .
Multiply the corrected moles by the molar mass: $m(\ce{Mg(OH)2}) = 0.01020\text{ mol} \times 58.32\text{ g/mol} = 0.594864\text{ g}$ .
Round to four significant figures to match the given data.

Final answer

$0.5949\text{ g}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice explicitly writing units with chemical formulas during stoichiometric conversion steps so units cancel visibly.
Perform a sanity check on purity percentages at the end of each analytical chemistry problem to catch mass values larger than the original sample.

OverviewACEpath Teacher's Diagnosis

03 The student adds the titrated excess moles to the initial acid amount instead of subtracting it to find the reacted moles.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$n(\text{total } \ce{HCl}) = 0.05000 \times 0.5000 = 0.02500\text{ mol}$
Step 2
$n(\ce{NaOH}) = 0.01840 \times 0.2500 = 0.004600\text{ mol}$
Step 3
$n(\text{reacted } \ce{HCl}) = 0.02500 + 0.004600 = 0.02960\text{ mol}$
Correct
$n(\ce{HCl})_{\text{reacted}} = n(\ce{HCl})_{\text{total}} - n(\ce{HCl})_{\text{excess}} = 0.02500\text{ mol} - 0.004600\text{ mol} = 0.02040\text{ mol}$
In a back-titration, the acid neutralized by $\ce{NaOH}$ is the leftover portion from the initial amount. Therefore, you must subtract the excess moles from the total initial moles ( $0.02500 - 0.004600$ ), rather than adding them.
Step 4
$n(\ce{Mg(OH)2}) = 0.02960 / 2 = 0.01480\text{ mol}$
Correct
$n(\ce{Mg(OH)2}) = \frac{0.02040\text{ mol}}{2} = 0.01020\text{ mol}$
You correctly applied the $1:2$ stoichiometric ratio from $\ce{Mg(OH)2 + 2HCl -> MgCl2 + 2H2O}$ , but the numerical value was carried forward from the addition error in Step 3.
Step 5
$m(\ce{Mg(OH)2}) = 0.01480 \times 58.32 = 0.8631\text{ g}$
Correct
$m(\ce{Mg(OH)2}) = 0.01020\text{ mol} \times 58.32\text{ g/mol} = 0.5949\text{ g}$
Your conversion from moles to grams using molar mass was methodologically sound, but the final numerical result is affected by the earlier error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly calculated the moles of both initial acid and titrant using $n = C \cdot V$ .

You recognized and applied the $1:2$ stoichiometric ratio between $\ce{Mg(OH)2}$ and $\ce{HCl}$ .

You used the correct molar mass of $\ce{Mg(OH)2}$ to convert moles to grams.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Back-Titration Mole Balance Error

You added the excess acid titrated by $\ce{NaOH}$ to the total initial acid instead of subtracting it.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the mole conservation relationship in a back-titration: total moles added represents the whole, while reacted moles and excess moles are the two complementary parts that sum to that total.

REASONING PATH

Your ACEpath teacher’s suggested approach

Back-Titration Part-Whole Conservation: $n_{\text{total}} = n_{\text{reacted with sample}} + n_{\text{excess back-titrated}}$
Rearrange to solve for the reacted acid: $n_{\text{reacted}} = n_{\text{total}} - n_{\text{excess}}$ .
Physical Reality Check: The moles of acid consumed by the sample cannot exceed the total initial moles of acid introduced.
Always verify that $n_{\text{reacted}} < n_{\text{total}}$ before proceeding to stoichiometric ratios.

CORRECTED PATH

ACEpath Teacher's Demonstration

Subtract the excess acid titrated by $\ce{NaOH}$ from the initial total acid: $n(\ce{HCl})_{\text{reacted}} = 0.02500\text{ mol} - 0.004600\text{ mol} = 0.02040\text{ mol}$ .
Divide by $2$ to find moles of $\ce{Mg(OH)2}$ : $n(\ce{Mg(OH)2}) = \frac{0.02040\text{ mol}}{2} = 0.01020\text{ mol}$ .
Multiply by the molar mass: $m(\ce{Mg(OH)2}) = 0.01020\text{ mol} \times 58.32\text{ g/mol} = 0.594864\text{ g} \approx 0.5949\text{ g}$ .

Final answer

$0.5949\text{ g}$

NEXT STEPS

Final Takeaways from Your Teacher

Review the basic concept of a back-titration as a two-stage process where total reagent equals reacted reagent plus unreacted leftover.
Write out the explicit word equation ' $n(\text{reacted}) = n(\text{total}) - n(\text{excess})$ ' before substituting numbers into the mole balance.
Complete the practice problems below to master back-titration setups.

Lock it in

Question

How do you relate the moles of a dibasic hydroxide $\ce{M(OH)2}$ to the moles of monoprotic acid $\ce{HCl}$ neutralized?

Answer

Since $\ce{M(OH)2 + 2HCl -> MCl2 + 2H2O}$ , each mole of base reacts with $2\text{ mol}$ of $\ce{HCl}$ , so $n(\ce{M(OH)2}) = \frac{1}{2} n(\ce{HCl})$ .

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student ignores the 1:2 stoichiometric ratio between Mg(OH)2 and HCl, assuming a 1:1 ratio instead.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

02 The student multiplies by 2 instead of dividing by 2 when converting moles of HCl to moles of Mg(OH)2.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

03 The student adds the titrated excess moles to the initial acid amount instead of subtracting it to find the reacted moles.

Sample Student Work

Step-by-Step Review

WHAT YOU DID WELL

THE ERROR

WHY IT HAPPENED

REASONING PATH

CORRECTED PATH

NEXT STEPS

Lock it in

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