In a titration of a weak acid, the end point was reached after adding 16.2 mL of base. Calculations showed that the equivalence point was reached after adding 15.6 mL of base.

How many mL of base was required to reach the point at which pH = pKa?

Question

In a titration of a weak acid, the end point was reached after adding 16.2 mL of base. Calculations showed that the equivalence point was reached after adding 15.6 mL of base.

How many mL of base was required to reach the point at which pH = pKa?

BlackTom AI · Accepted Answer

To locate the pH = pKa point in a titration of a weak acid with strong base, recall the half-equivalence principle: at the half-neutralization point, the amount of base added equals half of the amount needed to reach the equivalence point, and the pH equals the pKa of the weak acid.

Option 1: 16.2 mL. This volume corresponds to the endpoint and, for a weak acid–strong base titration, is not the half-equivalence point. Therefore, it does not give pH = pKa.

Option 2: 7.80 mL. Since the equivalence point occurs at 15.6 mL of base, half of that volume is 15.6 ÷ 2 = 7.8 mL. This is exactly the volume where the solution has equal moles of acid and its conjugate base, yielding pH = pKa by the Henderson–Hasselbalch framework.

Option 3: 8.10 mL. This value is slightly more than the half-equivalence volume. At any volume greater than half the equivalence volume, the pH would be higher than pKa, so this option does not satisfy pH = pKa.

Option 4: 15.6 mL. This is the equivalence point volume, where stoichiometric amounts have been mixed. At equivalence for a weak acid, the pH is typically > 7 and not equal to pKa. Hence this is not the pH = pKa point.

In summary, the only choice that aligns with pH = pKa for a weak acid titrated with a strong base is the half-equivalence volume, which is 7.80 mL. The other options correspond to the endpoint, a point slightly past half-equivalence, or the equivalence point, none of which satisfies pH = pKa.

CHEM 110 Lecture Quiz Week 10

In a titration of a weak acid, the end point was reached after adding 16.2 mL of base. Calculations showed that the equivalence point was reached after adding 15.6 mL of base. How many mL of base was required to reach the point at which pH = pKa?

View Explanation

Log in for full answers

Similar Questions

The given graph is a titration curve obtained for an acid base titration. How many protons are released from the acid used in this titration ? (Hint:asking whether monoprotic, diprotic...etc)

Which of the following is true for weak acid + strong base titration?

The given graph is a titration curve obtained for an acid base titration. How many protons are released from the acid used in this titration ? (Hint:asking whether monoprotic, diprotic...etc)

Titrate 50.0 mL of 0.100 M acetic acid with 0.100 M NaOH. What is the pH after the following titrant additions: 40.0 mL?

A 50.00 mL sample of 0.0950 M acetic acid (Ka = 1.8 × 10−5) is being titrated with 0.0848 M NaOH. What is the pH after 28.00 mL of NaOH has been added?

Which of the following indicator(s) would be most suitable for the titration of acetic acid (pKa = 4.74) with NaOH?

Question text Calculate the volume of NaOH needed to reach the equivalence point. Consider the titration of 50.0 mL of 0.100 M acetic acid, CH3COOH, with 0.200 M NaOH. Reaction equation: CH3COOH + NaOH → CH3COONa + H2O Volume of NaOH solution required (1 mark)= Answer 1 Question 44[input] mL (I:1)

An unknown solution of PhCOOH of 15.0 mL is titrated against a solution of 0.057 mol L-1 KOH. 12.7 mL of the base solution is required to reach the equivalence point. Calculate the concentration of the PhCOOH solution.

More Practical Tools for Students Powered by AI Study Helper

Homework AI Solver

Stylized AI Paper Writer

Plagiarism Checker Assistant

Citation AI Academic Writing Tool

In-Class Translation Assistant

AI Note Generator

AI Quiz Answers

Past Exam Questions from University Test Bank

Smart Practice Assistant

Adaptive Practice

Making Your Study Simpler