During an analytical chemistry back-titration experiment, the percent purity $(\%P)$ of an unknown solid acid sample of total mass $W_{\text{sample}}$ (in grams) is determined using the relation $\%P = \frac{(V_A M_A - V_B M_B) \times M_{\text{analyte}}}{10 \times W_{\text{sample}}}$, where $V_A$ and $V_B$ are volumes (in $\text{mL}$) of acid and base titrants, $M_A$ and $M_B$ are their respective molarities, and $M_{\text{analyte}}$ is the molar mass of the analyte. In a quality control protocol where the target percent purity $\%P$ is fixed, solve this equation explicitly for the required volume of back-titrant base $V

During an analytical chemistry back-titration experiment, the percent purity $(\%P)$ of an unknown solid acid sample of total mass $W_{\text{sample}}$ (in grams) is determined using the relation $\%P = \frac{(V_A M_A - V_B M_B) \times M_{\text{analyte}}}{10 \times W_{\text{sample}}}$ , where $V_A$ and $V_B$ are volumes (in $\text{mL}$ ) of acid and base titrants, $M_A$ and $M_B$ are their respective molarities, and $M_{\text{analyte}}$ is the molar mass of the analyte. In a quality control protocol where the target percent purity $\%P$ is fixed, solve this equation explicitly for the required volume of back-titrant base $V_B$ .

Final answer

$V_B = \frac{V_A M_A M_{\text{analyte}} - 10 (\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$

Steps

To solve for the target variable $V_B$ , treat all other variables as known constants and isolate $V_B$ using standard inverse algebraic operations.

Step 1: Clear the denominator Multiply both sides of the equation by the denominator, $10 \cdot W_{\text{sample}}$ : $10(\%P) W_{\text{sample}} = (V_A M_A - V_B M_B) M_{\text{analyte}}$

Step 2: Expand the right side Distribute the molar mass $M_{\text{analyte}}$ across the terms inside the parentheses: $10(\%P) W_{\text{sample}} = V_A M_A M_{\text{analyte}} - V_B M_B M_{\text{analyte}}$

Step 3: Isolate the term containing $V_B$ The term involving $V_B$ has a negative sign. Add $V_B M_B M_{\text{analyte}}$ to both sides and subtract $10(\%P) W_{\text{sample}}$ from both sides: $V_B M_B M_{\text{analyte}} = V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}$

Step 4: Solve for $V_B$ Divide both sides by the product $M_B M_{\text{analyte}}$ (assuming $M_B \neq 0$ and $M_{\text{analyte}} \neq 0$ ): $V_B = \frac{V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$

Common Pitfall: When dividing by $M_B M_{\text{analyte}}$ , make sure both terms in the numerator are included over the single denominator, rather than dividing only the adjacent term.

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to divide the left-hand side by M_analyte before subtracting V_A M_A, treating M_analyte as if it only multiplied the numerator or was already canceled.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$10 \%P W_{\text{sample}} = V_A M_A - V_B M_B$
Correct
$10 (\%P) W_{\text{sample}} = (V_A M_A - V_B M_B) M_{\text{analyte}}$
When multiplying both sides by the denominator $10 W_{\text{sample}}$ , the factor $M_{\text{analyte}}$ still multiplies the numerator $(V_A M_A - V_B M_B)$ and must be retained.
Step 2
$V_B M_B = V_A M_A - 10 \%P W_{\text{sample}}$
Correct
$V_B M_B M_{\text{analyte}} = V_A M_A M_{\text{analyte}} - 10 (\%P) W_{\text{sample}}$
You correctly rearranged terms according to your equation from Step 1, but the line inherits the missing factor $M_{\text{analyte}}$ .
Step 3
$V_B = \frac{V_A M_A - 10 \%P W_{\text{sample}}}{M_B}$
Correct
$V_B = \frac{V_A M_A M_{\text{analyte}} - 10 (\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$
You accurately divided across by the coefficient $M_B$ , but the final expression is missing $M_{\text{analyte}}$ due to the earlier omission.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared the denominator by multiplying both sides by $10 W_{\text{sample}}$ .

You isolated the term containing the target variable $V_B$ with sound addition and subtraction properties.

You correctly carried out the final division to isolate $V_B$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted numerator factor during clearing

You accidentally dropped the factor $M_{\text{analyte}}$ when clearing the fraction's denominator.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating a fraction of the form $\frac{(A - B)C}{D}$ as $\frac{A - B}{D}$ when clearing $D$ , losing track of the remaining multiplier $C$ in the numerator.

REASONING PATH

Your ACEpath teacher’s suggested approach

Account for every numerator factor: Multiplying $\frac{(A - B)C}{D}$ by $D$ cancels only the denominator $D$ , leaving $(A - B)C$ .
Circle or group all factors in the numerator before clearing the denominator so that none are left behind.

CORRECTED PATH

ACEpath Teacher's Demonstration

Multiply both sides by $10 W_{\text{sample}}$ while keeping all numerator factors: $10(\%P) W_{\text{sample}} = (V_A M_A - V_B M_B) M_{\text{analyte}}$ .
Distribute $M_{\text{analyte}}$ on the right-hand side: $10(\%P) W_{\text{sample}} = V_A M_A M_{\text{analyte}} - V_B M_B M_{\text{analyte}}$ .
Isolate the term with $V_B$ : $V_B M_B M_{\text{analyte}} = V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}$ .
Divide both sides by $M_B M_{\text{analyte}}$ to solve for $V_B$ .

Final answer

$V_B = \frac{V_A M_A M_{\text{analyte}} - 10 (\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$

NEXT STEPS

Final Takeaways from Your Teacher

Review clearing fractions where the numerator is a product of multiple factors.
Practice distributing external factors before attempting to isolate literal terms.
Verify your final rearranged equation by checking units or dimensional consistency.

OverviewACEpath Teacher's Diagnosis

02 The student makes a sign error when isolating the negative term -V_B M_B, resulting in the terms in the numerator being subtracted in the reverse order.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$10(\%P) W_{\text{sample}} = (V_A M_A - V_B M_B) M_{\text{analyte}} \implies \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}} = V_A M_A - V_B M_B$
Step 2
$V_B M_B = \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}} - V_A M_A$
Correct
$-V_B M_B = \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}} - V_A M_A \implies V_B M_B = V_A M_A - \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}}$
When subtracting $V_A M_A$ from both sides, the term remaining on the right side retains its negative sign, giving $-V_B M_B$ . Multiplying or dividing by $-1$ reverses the terms to $V_A M_A - \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}}$ .
Step 3
$V_B = \frac{\frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}} - V_A M_A}{M_B}$
Correct
$V_B = \frac{V_A M_A - \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}}}{M_B} = \frac{V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$
You carried out the division by $M_B$ correctly according to your previous line, but the result inherits the sign error from Step 2.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared the fraction by multiplying both sides by the denominator $10 W_{\text{sample}}$ .

You accurately divided both sides by $M_{\text{analyte}}$ to isolate the bracketed terms.

You correctly divided by the coefficient $M_B$ in the final step.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Dropped Negative Sign During Variable Isolation

When you subtracted $V_A M_A$ from both sides, you wrote positive $V_B M_B$ instead of keeping the minus sign on $-V_B M_B$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Treating a subtracted term $A - Bx$ as simply $Bx$ after subtracting $A$ , rather than tracking the negative coefficient attached to the term being isolated.

REASONING PATH

Your ACEpath teacher’s suggested approach

Track the Negative Sign: In an expression $A - B$ , subtracting $A$ leaves $-B$ , not $+B$ .
Always circle or highlight the minus sign attached to a term before moving terms across the equals sign, or alternatively add the negative term to the other side first.
Alternative Move Strategy: If the variable to isolate has a negative sign, add it to both sides first: $C = A - Bx \implies C + Bx = A \implies Bx = A - C$ .
Add the negative term to both sides early to avoid sign confusion.

CORRECTED PATH

ACEpath Teacher's Demonstration

Start with the correct intermediate equation: $\frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}} = V_A M_A - V_B M_B$
Add $V_B M_B$ to both sides and subtract $\frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}}$ : $V_B M_B = V_A M_A - \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}}$
Divide both sides by $M_B$ : $V_B = \frac{V_A M_A - \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}}}{M_B}$
Simplify the compound fraction by multiplying the numerator and denominator by $M_{\text{analyte}}$ : $V_B = \frac{V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$

Final answer

$V_B = \frac{V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice isolating literal terms that have negative coefficients by adding them to the opposite side first.
Review how to clear compound fractions into a single rational expression.

OverviewACEpath Teacher's Diagnosis

03 The student divides by M_B before isolating the V_B term but fails to divide all terms on the right-hand side, erroneously isolating V_B while leaving other terms unscaled.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\frac{(10\ \%P\ W_{\text{sample}})}{M_{\text{analyte}}} = V_A M_A - V_B M_B$
Step 2
$\frac{(10\ \%P\ W_{\text{sample}})}{(M_{\text{analyte}} M_B)} = V_A M_A - V_B$
Correct
$\frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}} M_B} = \frac{V_A M_A}{M_B} - V_B$
When dividing both sides of an equation by a quantity such as $M_B$ , every term on each side must be divided. You divided the left side and the $V_B M_B$ term by $M_B$ , but you left $V_A M_A$ undivided.
Step 3
$V_B = V_A M_A - \frac{(10\ \%P\ W_{\text{sample}})}{M_{\text{analyte}} M_B}$
Correct
$V_B = \frac{V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$
You rearranged the terms consistently from your previous line, but because the equation was already missing the division of $V_A M_A$ by $M_B$ , the resulting expression for $V_B$ is incorrect.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly cleared the fraction by multiplying both sides by the denominator $10 W_{\text{sample}}$ .

You correctly isolated the binomial term $V_A M_A - V_B M_B$ by dividing by $M_{\text{analyte}}$ .

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Incomplete Division Across Terms

Dividing only the targeted variable term by its coefficient instead of dividing every term on that side of the equation.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Dividing by a coefficient before fully isolating the target term. When dividing an entire side of an equation by a factor, the division must distribute over every additive term: $\frac{A - B}{C} = \frac{A}{C} - \frac{B}{C}$ . It is much safer to isolate the term containing the variable first using addition/subtraction before dividing by its coefficient.

REASONING PATH

Your ACEpath teacher’s suggested approach

Isolate the target term first: Always isolate the term containing the desired variable using addition or subtraction before dividing by its coefficient.
Add $V_B M_B$ to both sides and subtract the left-hand expression so that the $V_B$ term stands entirely alone.
Distribute division across all terms: If dividing an expression with multiple terms, apply division to every single term: $(X - Y) / Z = X/Z - Y/Z$ .
Ensure no term is left unscaled when dividing both sides by a constant or variable factor.

CORRECTED PATH

ACEpath Teacher's Demonstration

Isolate the term containing $V_B$ using addition and subtraction: $V_B M_B = V_A M_A - \frac{10(\%P) W_{\text{sample}}}{M_{\text{analyte}}}$ .
Combine the terms on the right-hand side over a common denominator: $V_B M_B = \frac{V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}}{M_{\text{analyte}}}$ .
Divide both sides by $M_B$ : $V_B = \frac{V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$ .

Final answer

$V_B = \frac{V_A M_A M_{\text{analyte}} - 10(\%P) W_{\text{sample}}}{M_B M_{\text{analyte}}}$

NEXT STEPS

Final Takeaways from Your Teacher

Practice isolating literal terms using addition/subtraction before dividing by coefficients.
When performing division across a polynomial or binomial, write the entire side over a single fraction bar to avoid missing terms.
Check algebraic operations by substituting simple non-zero test numbers for each variable.

Lock it in

Question

What remains on the right-hand side when you multiply $y = \frac{(a - b)c}{d}$ by $d$ ?

Answer

$(a - b)c$ . Multiplying by the denominator cancels only $d$ , leaving the entire numerator unchanged.

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to divide the left-hand side by M_analyte before subtracting V_A M_A, treating M_analyte as if it only multiplied the numerator or was already canceled.

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 makes a sign error when isolating the negative term -V_B M_B, resulting in the terms in the numerator being subtracted in the reverse order.

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 divides by M_B before isolating the V_B term but fails to divide all terms on the right-hand side, erroneously isolating V_B while leaving other terms unscaled.

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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