Multiple Choice
A 50.0 mL volume of 0.15 mol L⁻¹ HBr is titrated with 0.25 mol L⁻¹ KOH. Calculate the pH after the addition of 12.0 mL of KOH.
18. Aqueous Equilibrium
Titrations: Strong Acid-Strong Base
Struggling with General Chemistry?
Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
Consider the titration of a 25.0 mL sample of 0.115 M RbOH with 0.100 M HCl. Determine the pH at 5.0 mL of added acid.
A
11.85
B
12.30
C
2.15
D
7.00
Verified step by step guidance1
Calculate the initial moles of RbOH in the solution using the formula: \( \text{moles of RbOH} = \text{volume (L)} \times \text{molarity (M)} \). Convert 25.0 mL to liters by dividing by 1000.
Calculate the moles of HCl added using the formula: \( \text{moles of HCl} = \text{volume (L)} \times \text{molarity (M)} \). Convert 5.0 mL to liters by dividing by 1000.
Determine the moles of RbOH remaining after the reaction with HCl. Since RbOH and HCl react in a 1:1 ratio, subtract the moles of HCl from the initial moles of RbOH.
Calculate the concentration of RbOH remaining in the solution. The total volume of the solution is the sum of the initial volume of RbOH and the volume of HCl added. Convert this total volume to liters and use it to find the new concentration: \( \text{concentration} = \frac{\text{moles of RbOH remaining}}{\text{total volume (L)}} \).
Determine the pH of the solution. Since RbOH is a strong base, use the concentration of OH\(^-\) ions to find the pOH: \( \text{pOH} = -\log[\text{OH}^-] \). Then, calculate the pH using the relation: \( \text{pH} = 14 - \text{pOH} \).
1:09mWatch next
Master Strong Acid-Strong Base Titration with a bite sized video explanation from Jules
Start learningRelated Videos
Related Practice
Titrations: Strong Acid-Strong Base practice set
Problem sets built by lead tutorsExpert video explanations