Multiple Choice
Determine the pH of a 7.8×10⁻⁴ M Ba(OH)₂ solution.
17. Acid and Base Equilibrium
pH of Strong Acids and Bases
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
What is the pH of a solution after mixing 250 mL of 0.250 M HBr with 250 mL of 0.125 M NaOH?
A
1.00
B
7.00
C
13.00
D
2.00
Verified step by step guidance1
First, identify the type of reaction occurring. HBr is a strong acid and NaOH is a strong base, so they will undergo a neutralization reaction: \( \text{HBr} + \text{NaOH} \rightarrow \text{NaBr} + \text{H}_2\text{O} \).
Calculate the moles of HBr and NaOH present in the solution. Use the formula \( \text{moles} = \text{concentration} \times \text{volume} \). For HBr: \( 0.250 \text{ M} \times 0.250 \text{ L} = 0.0625 \text{ moles} \). For NaOH: \( 0.125 \text{ M} \times 0.250 \text{ L} = 0.03125 \text{ moles} \).
Determine the limiting reactant by comparing the moles of HBr and NaOH. Since NaOH has fewer moles (0.03125 moles), it is the limiting reactant and will be completely consumed in the reaction.
Calculate the moles of HBr remaining after the reaction. Subtract the moles of NaOH from the initial moles of HBr: \( 0.0625 \text{ moles} - 0.03125 \text{ moles} = 0.03125 \text{ moles} \) of HBr remain.
Find the concentration of the remaining HBr in the final solution. The total volume of the solution after mixing is \( 250 \text{ mL} + 250 \text{ mL} = 500 \text{ mL} = 0.500 \text{ L} \). The concentration of HBr is \( \frac{0.03125 \text{ moles}}{0.500 \text{ L}} = 0.0625 \text{ M} \). Finally, calculate the pH using the formula \( \text{pH} = -\log[\text{H}^+] \), where \([\text{H}^+] = 0.0625 \text{ M}\).
2:39mWatch next
Master pH of Strong Acids and Bases with a bite sized video explanation from Jules
Start learningRelated Videos
Related Practice
pH of Strong Acids and Bases practice set
Problem sets built by lead tutorsExpert video explanations