In an analytical chemistry laboratory, a student titrates a weak monoprotic acid with standard sodium hydroxide. The pH buffer region is considered optimal when the hydronium ion concentration $[\text{H}_3\text{O}^+]$ satisfies $1.80 \times 10^{-5}\,\text{M} \le [\text{H}_3\text{O}^+] \le 7.20 \times 10^{-5}\,\text{M}$. Using the relation $\text{pH} = -\log_{10}[\text{H}_3\text{O}^+]$, solve for the corresponding acceptable range of the solution's $\text{pH}$. Express the final boundary values rounded to two decimal places in interval notation.

In an analytical chemistry laboratory, a student titrates a weak monoprotic acid with standard sodium hydroxide. The pH buffer region is considered optimal when the hydronium ion concentration $[\text{H}_3\text{O}^+]$ satisfies $1.80 \times 10^{-5}\,\text{M} \le [\text{H}_3\text{O}^+] \le 7.20 \times 10^{-5}\,\text{M}$ . Using the relation $\text{pH} = -\log_{10}[\text{H}_3\text{O}^+]$ , solve for the corresponding acceptable range of the solution's $\text{pH}$ . Express the final boundary values rounded to two decimal places in interval notation.

Final answer

$[4.14, 4.74]$

Steps

Set up the compound inequality

To find the acceptable range for the solution's $\text{pH}$ , we apply the logarithm and the negative sign to the given compound inequality.

The optimal hydronium ion concentration is bounded by: $1.80 \times 10^{-5} \le [\ce{H3O+}] \le 7.20 \times 10^{-5}$

Apply the base-10 logarithm

Because the function $f(x) = \log_{10}(x)$ is strictly increasing for all $x > 0$ , taking the logarithm of each part preserves the direction of the inequality signs: $\log_{10}(1.80 \times 10^{-5}) \le \log_{10}[\ce{H3O+}] \le \log_{10}(7.20 \times 10^{-5})$

Using logarithm properties, $\log_{10}(a \cdot 10^b) = \log_{10}(a) + b$ :

For the lower concentration bound: $\log_{10}(1.80 \times 10^{-5}) = \log_{10}(1.80) - 5 \approx 0.2553 - 5 = -4.7447$
For the upper concentration bound: $\log_{10}(7.20 \times 10^{-5}) = \log_{10}(7.20) - 5 \approx 0.8573 - 5 = -4.1427$

Substituting these approximations yields: $-4.7447 \le \log_{10}[\ce{H3O+}] \le -4.1427$

Multiply by $-1$ to isolate $\text{pH}$

The definition of $\text{pH}$ is $\text{pH} = -\log_{10}[\ce{H3O+}]$ .

Multiplying an inequality across by a negative number reverses the inequality signs: $-(-4.7447) \ge -\log_{10}[\ce{H3O+}] \ge -(-4.1427)$ $4.7447 \ge \text{pH} \ge 4.1427$

Rewriting this from smallest to largest value: $4.1427 \le \text{pH} \le 4.7447$

Round and write in interval notation

Rounding the lower and upper bounds to two decimal places:

Lower bound: $4.14$
Upper bound: $4.74$

In interval notation, the closed range of optimal $\text{pH}$ values is: $[4.14, 4.74]$

Educational illustration for this question

Typical errors

OverviewACEpath Teacher's Diagnosis

01 The student forgets to reverse the inequality signs when applying the negative logarithm, writing the interval with the upper bound first.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Partially correct

Step 1
$\text{pH}$ 1 = -\log{10}(1.80 $\times$ 10^{-5}) = 4.74 $\implies$ $\text{pH}$ 2 = -\log{10}(7.20 $\times$ 10^{-5}) = 4.14
Step 2
1.80 $\times$ 10^{-5} $\le$ [ $\text{H}$ 3 $\text{O}$ ^+] $\le$ 7.20 $\times$ 10^{-5} $\implies$ 4.74 $\le$ $\text{pH}$ $\le$ 4.14
Correct
-\log_{10}(1.80 $\times$ 10^{-5}) $\ge$ -\log_{10}[ $\text{H}$ 3 $\text{O}$ ^+] $\ge$ -\log{10}(7.20 $\times$ 10^{-5}) $\implies$ 4.14 $\le$ $\text{pH}$ $\le$ 4.74
Because the function $f(x) = -\log_{10}(x)$ is strictly decreasing (multiplying by $-1$ reverses an inequality), the inequality signs must flip. The higher hydronium concentration corresponds to the lower pH.
Step 3
[4.74, 4.14]
Correct
[4.14, 4.74]
This step inherited the inverted bounds from the previous step. In interval notation $[a, b]$ , the smaller lower bound must always be written first.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly used the formula $\text{pH} = -\log_{10}[\text{H}_3\text{O}^+]$ .

You computed and rounded both boundary pH values accurately to two decimal places.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Failure to reverse inequality direction under a decreasing transformation

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

You substituted the transformed boundary values directly into the original inequality positions without accounting for the fact that multiplying by a negative number (or applying a strictly decreasing function) reverses the order of the inequalities.

REASONING PATH

Your ACEpath teacher’s suggested approach

Inequality Reversal Rule: Whenever you multiply or divide an inequality by a negative number, or apply a strictly decreasing function, the inequality signs reverse direction ( $a \le x \le b \implies -a \ge -x \ge -b$ ).
Always check that the lower numerical bound is placed on the left: if $4.74 \ge \text{pH} \ge 4.14$ , rewrite it in standard order as $4.14 \le \text{pH} \le 4.74$ .
Interval Notation Ordering: Interval notation $[a, b]$ strictly requires that $a \le b$ .
Before writing down your interval, confirm that the first number is less than or equal to the second number.

CORRECTED PATH

ACEpath Teacher's Demonstration

Start with the compound inequality: $1.80 \times 10^{-5} \le [\text{H}_3\text{O}^+] \le 7.20 \times 10^{-5}$
Apply the strictly increasing base-10 logarithm: $\log_{10}(1.80 \times 10^{-5}) \le \log_{10}[\text{H}_3\text{O}^+] \le \log_{10}(7.20 \times 10^{-5})$ , which gives $-4.7447 \le \log_{10}[\text{H}_3\text{O}^+] \le -4.1427$
Multiply across by $-1$ and reverse the inequality signs: $4.7447 \ge -\log_{10}[\text{H}_3\text{O}^+] \ge 4.1427$ , meaning $4.1427 \le \text{pH} \le 4.7447$
Round the boundaries to two decimal places and express in interval notation: $[4.14, 4.74]$

Final answer

[4.14, 4.74]

NEXT STEPS

Final Takeaways from Your Teacher

Practice solving compound inequalities where multiplying across by $-1$ flips the inequality symbols.
Remember that higher acidity (higher $[\text{H}_3\text{O}^+]$ ) results in a lower pH value.
Always verify that any interval $[a, b]$ has its smaller value on the left.

OverviewACEpath Teacher's Diagnosis

02 The student omits the negative sign in the pH definition and calculates log₁₀[H₃O⁺], leading to negative pH values.

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\text{pH} = \log_{10}[\ce{H3O+}]$
Correct
$\text{pH} = -\log_{10}[\ce{H3O+}]$
The problem defines $\text{pH} = -\log_{10}[\ce{H3O+}]$ , but you omitted the negative sign in your definition.
Step 2
\log_{10}(1.80 $\times$ 10^{-5}) = -4.74 $\implies$ \log_{10}(7.20 $\times$ 10^{-5}) = -4.14
Step 3
-4.74 $\le$ $\text{pH}$ $\le$ -4.14
Correct
-(-4.74) $\ge$ -\log_{10}[ $\ce{H3O+}$ ] $\ge$ -(-4.14) $\implies$ 4.14 $\le$ $\text{pH}$ $\le$ 4.74_
Because the negative sign was omitted in your definition, you substituted $\text{pH}$ directly into the inequality without multiplying by $-1$ , which also means the inequality directions were never reversed.
Step 4
[-4.74, -4.14]
Correct
[4.14, 4.74]
Your interval notation correctly matches your compound inequality in Step 3, but the values are incorrect due to the initial formula error.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly calculated the base-10 logarithms of both scientific notation concentration values.

You expressed your compound inequality correctly in closed interval notation.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Omitted Negative Sign in Formula / Inequality Direction Flip

You dropped the leading negative sign in $\text{pH} = -\log_{10}[\ce{H3O+}]$ , writing $\text{pH} = \log_{10}[\ce{H3O+}]$ instead.

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Overlooking the negative factor in the definition of $\text{pH}$ meant you did not multiply the compound inequality by $-1$ , missing both the sign change to positive $\text{pH}$ values and the required reversal of inequality signs.

REASONING PATH

Your ACEpath teacher’s suggested approach

Check the full formula before computing: Always copy the exact formula from the prompt: $\text{pH} = -\log_{10}[\ce{H3O+}]$ .
Verify that every sign and operator in the prompt's formula matches your written setup before substituting values.
Reverse inequality signs when multiplying by a negative: Multiplying or dividing an inequality by a negative number reverses the direction of all inequality symbols: if $a \le x \le b$ , then $-b \le -x \le -a$ .
When isolating $-x$ or $-\log_{10}[\ce{H3O+}]$ , flip $\le$ to $\ge$ and rewrite the inequality in standard order from smaller to larger.

CORRECTED PATH

ACEpath Teacher's Demonstration

Start with the logarithm inequality: $-4.7447 \le \log_{10}[\ce{H3O+}] \le -4.1427$ .
Multiply the entire compound inequality by $-1$ and reverse the inequality symbols: $4.7447 \ge -\log_{10}[\ce{H3O+}] \ge 4.1427$ .
Replace $-\log_{10}[\ce{H3O+}]$ with $\text{pH}$ : $4.1427 \le \text{pH} \le 4.7447$ .
Round each boundary to two decimal places: $4.14 \le \text{pH} \le 4.74$ .

Final answer

[4.14, 4.74]

NEXT STEPS

Final Takeaways from Your Teacher

Review the effect of multiplying or dividing an inequality by $-1$ on the direction of inequality symbols.
Double-check that chemical $\text{pH}$ values in dilute aqueous solutions are typically positive numbers between $0$ and $14$ as a sanity check.

OverviewACEpath Teacher's Diagnosis

03 The student misapplies the logarithm product rule by subtracting rather than adding inside the parenthesis, computing -(log₁₀(c) - 5) incorrectly as -(log₁₀(c) + 5).

Sample Student Work

Reviewed by ACEpath Teacher

Step-by-Step Review

Walk Through It with Your Teacher

Incorrect

Step 1
$\text{pH}$ = -(\log_{10}(1.80) + \log_{10}(10^{-5})) $\implies$ -(0.26 + 5) = -5.26
Correct
$\text{pH}$ = -(\log_{10}(1.80) + \log_{10}(10^{-5})) = -(0.26 - 5) = 4.74
By the power property of logarithms, $\log_{10}(10^k) = k$ , so $\log_{10}(10^{-5}) = -5$ . Replacing $\log_{10}(10^{-5})$ with $+5$ gave $-(0.26 + 5) = -5.26$ instead of $-(0.26 - 5) = 4.74$ .
Step 2
$\text{pH}$ = -(\log_{10}(7.20) + \log_{10}(10^{-5})) $\implies$ -(0.86 + 5) = -5.86
Correct
$\text{pH}$ = -(\log_{10}(7.20) + \log_{10}(10^{-5})) = -(0.86 - 5) = 4.14
You repeated the same sign mistake here: $\log_{10}(10^{-5})$ is $-5$ , giving $-(0.86 - 5) = 4.14$ rather than $-(0.86 + 5) = -5.86$ .
Step 3
[-5.86, -5.26]
Correct
[4.14, 4.74]
Because both calculated boundary values carried the sign error from the previous steps, the resulting interval was incorrect.

WHAT YOU DID WELL

Compliments from Your Teacher

You correctly identified the formula $\text{pH} = -\log_{10}[\ce{H3O+}]$ .

You correctly applied the product rule for logarithms, writing $\log_{10}(c \cdot 10^{-5}) = \log_{10}(c) + \log_{10}(10^{-5})$ .

You correctly ordered your calculated endpoints from smallest to largest in interval notation.

THE ERROR

Where Your ACEpath Teacher Spotted the Mistake

Logarithm of Negative Power of Ten

You dropped the negative sign on the exponent when evaluating $\log_{10}(10^{-5})$ , writing $+5$ instead of $-5$ .

WHY IT HAPPENED

ACEpath Teacher's Diagnosis

Confusing the exponent $k$ in $\log_{10}(10^k)$ with its absolute value $|k|$ , forgetting that negative powers of 10 yield negative logarithms.

REASONING PATH

Your ACEpath teacher’s suggested approach

Logarithm of a Power of Ten: For any real exponent $k$ , $\log_{10}(10^k) = k$ .
Always keep the sign of the exponent: $\log_{10}(10^{-n}) = -n$ , never $+n$ .
Order Reversal Under Negation: If $a \le x \le b$ , then $-b \le -x \le -a$ .
When taking $-\log_{10}$ , the upper concentration bound produces the lower $\text{pH}$ bound.

CORRECTED PATH

ACEpath Teacher's Demonstration

Evaluate the lower concentration boundary correctly: $\text{pH} = -(\log_{10}(1.80) + (-5)) \approx -(0.2553 - 5) = 4.7447$ .
Evaluate the upper concentration boundary correctly: $\text{pH} = -(\log_{10}(7.20) + (-5)) \approx -(0.8573 - 5) = 4.1427$ .
Because multiplying by $-1$ reverses the inequality directions, $4.1427 \le \text{pH} \le 4.7447$ .

Final answer

[4.14, 4.74]

NEXT STEPS

Final Takeaways from Your Teacher

Practice evaluating base-10 logarithms of scientific notation numbers like $a \times 10^{-n}$ to make the sign change automatic.
Perform a physical reality check: typical aqueous buffer solutions have positive $\text{pH}$ values (normally between 0 and 14), so a negative $\text{pH}$ should immediately prompt you to recheck your signs.

Lock it in

Question

What happens to the direction of inequality signs when every term is multiplied or divided by a negative number?

Answer

The inequality signs reverse direction ( $<$ becomes $>$ , and $\le$ becomes $\ge$ ).

Resources

Answer and steps

Final answer

Steps

Typical errors

01 The student forgets to reverse the inequality signs when applying the negative logarithm, writing the interval with the upper bound first.

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 omits the negative sign in the pH definition and calculates log₁₀[H₃O⁺], leading to negative pH values.

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 misapplies the logarithm product rule by subtracting rather than adding inside the parenthesis, computing -(log₁₀(c) - 5) incorrectly as -(log₁₀(c) + 5).

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

Similar question