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?

Question

BlackTom AI · Accepted Answer

We begin by identifying the reaction and calculating the quantities involved. The acetic acid  (CH3COOH) has initial moles = 0.0500 L × 0.0950 M = 0.00475 mol. The NaOH added has moles = 0.02800 L × 0.0848 M = 0.0023744 mol. Since the base amount is less than the acid, we form a buffer consisting of the weak acid and its conjugate base (acetate). The remaining acid after neutralization is 0.00475 − 0.0023744 = 0.0023756 mol, and the produced conjugate base (acetate) is 0.0023744 mol.

The total volume after addition is 50.0 mL + 28.0 mL = 78.0 mL = 0.078 L. Therefore, the concentrations in the buffer are: [HA] = 0.0023756 / 0.078 ≈ 0.03044 M and [A−] = 0.0023744 / 0.078 ≈ 0.03044 M. These values are essentially equal, indicating a nearly equimolar buffer.

Using the Henderson–Hasselbalch equation for a buffer: pH = pKa + log([A−]/[HA]). The Ka for acetic acid is 1.8 × 10−5, so pKa ≈ −log10(1.8×10−5) ≈ 4.7447. With [A−] ≈ [HA], the ratio [A−]/[HA] ≈ 1, and log(1) = 0. Thus pH ≈ pKa ≈ 4.745, which rounds to about 4.74.

Now, evaluate each answer option:
- 3.18: This is far below the pKa region and would occur if the solution were dominated by the weak acid with little conjugate base present. The calculated [A−]/[HA] ratio near 1 rules this out, so this is incorrect.
- 3.06: Similar to 3.18, this value lies well into a more acidic pH region than our buffer calculation supports; the buffer composition does not support such a low pH here, making this option incorrect.
- 4.44: This value is lower than the expected pH for a near-equimolar acetic acid/acetate buffer at this Ka; while in the correct range, it does not align with the calculated pH and would imply a significantly larger imbalance between [A−] and [HA], so this is misleading.
- 5.04: This would correspond to a much more basic solution or a buffer with substantially more conjugate base than acid, which is not the case since moles of acid and base consumed were nearly equal; thus this option is incorrect.
- 4.74: This option matches the calculated pH from the Henderson–Hasselbalch treatment for a near-equimolar acetic acid/acetate buffer formed after partial neutralization, making it the correct choice.

In summary, the calculation shows the system behaves as a weak acid–conjugate base buffer with nearly equal concentrations of CH3COOH and CH3COO− after adding 28.0 mL of NaOH, yielding a pH approximately equal to pKa, which is about 4.74. The other values arise from scenarios inconsistent with the buffer composition at this equivalence stage.

CHM-1020-LD01.2025SU Practice Exam II

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?

查看解析

登录即可查看完整答案

类似问题

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?

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)

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?

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.

更多留学生实用工具

考试浏览器助手

风格化写作助手

论文查重助手

文献引用助手

课堂转译助手

课堂笔记助手

Quiz搜索助手

学校历年真题

智能刷题助手

智能匹配练习

希望你的学习变得更简单