Titration Curves & Indicators practice questions

Question 1

A student titrates a $20.00 \text{ mL}$ sample of an $\text{HCl(aq)}$ solution of unknown concentration with a standardized $0.100 \text{ M NaOH(aq)}$ solution using phenolphthalein as an indicator. The student records the buret readings for three trials in the table below.

| Trial | Volume of $\text{HCl(aq)}$ ($\text{mL}$) | Initial buret reading ($\text{mL}$) | Final buret reading ($\text{mL}$) |
|-------|-------------------------------------------|--------------------------------------|------------------------------------|
| 1     | 20.00                                     | 0.50                                 | 10.50                              |
| 2     | 20.00                                     | 10.50                                | 20.50                              |
| 3     | 20.00                                     | 20.50                                | 30.50                              |

Based on the titration data, what is the molar concentration of the $\text{HCl(aq)}$ solution?

Accepted Answer

$0.0500 \text{ M}$

Answer

$0.100 \text{ M}$

Answer

$0.200 \text{ M}$

Answer

$0.400 \text{ M}$

Question 2

A student titrates an acidic solution of oxalic acid, $\text{H}_2\text{C}_2\text{O}_4\text{(aq)}$, with a standardized potassium permanganate solution, $\text{KMnO}_4\text{(aq)}$. The balanced net-ionic equation for the redox titration reaction is shown below:

$$2\text{MnO}_4^-\text{(aq)} + 5\text{H}_2\text{C}_2\text{O}_4\text{(aq)} + 6\text{H}^+\text{(aq)} \rightarrow 2\text{Mn}^{2+}\text{(aq)} + 10\text{CO}_2\text{(g)} + 8\text{H}_2\text{O(l)}$$

Which species acts as the reducing agent in this reaction, and what is the change in the oxidation number of the atom being oxidized?

Accepted Answer

$\text{H}_2\text{C}_2\text{O}_4\text{(aq)}$, in which the oxidation number of $\text{C}$ increases from $+3$ to $+4$

Answer

$\text{MnO}_4^-\text{(aq)}$, in which the oxidation number of $\text{Mn}$ decreases from $+7$ to $+2$

Answer

$\text{H}^+\text{(aq)}$, in which the oxidation number of $\text{H}$ increases from $+1$ to $+2$

Answer

$\text{H}_2\text{C}_2\text{O}_4\text{(aq)}$, in which the oxidation number of $\text{C}$ increases from $+2$ to $+4$

Question 3

A student titrates a sample of an unknown monoprotic weak acid with standard $0.10 \text{ M NaOH(aq)}$ using a pH meter and generates the titration curve shown. The student plans to repeat the experiment using a chemical indicator to signal the endpoint instead of using a pH meter. Based on the titration curve, which of the following correctly identifies the volume of $\text{NaOH(aq)}$ required to reach the equivalence point and selects the most appropriate indicator for the titration?

Accepted Answer

Volume: $25.0 \text{ mL}$; Indicator: Phenolphthalein (pH range 8.2–10.0), because its color-change range coincides with the steep pH jump surrounding the equivalence point.

Answer

Volume: $12.5 \text{ mL}$; Indicator: Methyl red (pH range 4.4–6.2), because the equivalence point occurs where the pH changes most gradually in the buffer region.

Answer

Volume: $25.0 \text{ mL}$; Indicator: Methyl red (pH range 4.4–6.2), because its color-change range matches the initial pH range of the weak acid solution.

Answer

Volume: $50.0 \text{ mL}$; Indicator: Phenolphthalein (pH range 8.2–10.0), because the equivalence point is reached when the solution pH levels off at its maximum value.

Question 4

A student titrates a $50.0\text{ mL}$ sample of a weak monoprotic acid, $\text{HA(aq)}$, with $0.100\text{ M }\text{NaOH(aq)}$. The titration curve for the process is shown below. Based on the titration curve, which of the following correctly identifies the approximate $K_a$ of $\text{HA}$ and provides the correct justification?

Accepted Answer

$1.0 \times 10^{-5}$, because at $20.0\text{ mL}$ of added $\text{NaOH(aq)}$, $[\text{HA}] = [\text{A}^-]$ and $\text{pH} = p K_a = 5.0$.

Answer

$1.0 \times 10^{-5}$, because at $40.0\text{ mL}$ of added $\text{NaOH(aq)}$, $[\text{HA}] = [\text{A}^-]$ and $\text{pH} = p K_a = 5.0$.

Answer

$1.0 \times 10^{-9}$, because at $40.0\text{ mL}$ of added $\text{NaOH(aq)}$, the equivalence point is reached where $\text{pH} = p K_a = 9.0$.

Answer

$1.0 \times 10^{-3}$, because at $0.0\text{ mL}$ of added $\text{NaOH(aq)}$, the initial $\text{pH}$ of the acid solution equals its $p K_a$.

Question 5

A student titrates a sample of a weak diprotic acid, $\text{H}_2\text{A}$, with $0.10 \text{ M NaOH}$. The titration curve generated from the experimental data is shown below. Based on the titration curve, which of the following best identifies the value of $pK_{a2}$ for $\text{H}_2\text{A}$ and the reasoning used to determine it?

Accepted Answer

$7.2$, because it is the pH at the second half-equivalence point at $30.0 \text{ mL}$ of added $\text{NaOH}$

Answer

$3.5$, because it is the pH at the first half-equivalence point at $10.0 \text{ mL}$ of added $\text{NaOH}$

Answer

$6.0$, because it is the pH at the first equivalence point at $20.0 \text{ mL}$ of added $\text{NaOH}$

Answer

$10.0$, because it is the pH at the second equivalence point at $40.0 \text{ mL}$ of added $\text{NaOH}$

Question 6

A student titrates a 20.0 mL sample of a diprotic acid solution, $\text{H}_2\text{A(aq)}$, with a standardized $0.100 \text{ M NaOH(aq)}$ solution using phenolphthalein as an indicator. The net ionic equation for the reaction that occurs during the titration is shown below.

$$\text{H}_2\text{A(aq)} + 2\text{OH}^-\text{(aq)} \rightarrow \text{A}^{2-}\text{(aq)} + 2\text{H}_2\text{O(l)}$$

The initial buret reading of $0.100 \text{ M NaOH(aq)}$ is 5.00 mL. At the pale pink endpoint, the final buret reading is 45.00 mL. What is the molarity of the $\text{H}_2\text{A(aq)}$ sample?

Accepted Answer

$0.100 \text{ M}$

Answer

$0.0333 \text{ M}$

Answer

$0.0667 \text{ M}$

Answer

$0.200 \text{ M}$

Question 7

A student titrates a sample of a monoprotic acid, $\text{HA(aq)}$, with standardized $\text{NaOH(aq)}$. The reaction is represented by

$\text{HA(aq)}+\text{OH}^-\text{(aq)} \rightarrow \text{A}^-\text{(aq)}+\text{H}_2\text{O(l)}$

| Quantity | Value |
|---|---:|
| Volume of acid sample | $25.00\text{ mL}$ |
| Concentration of $\text{NaOH(aq)}$ | $0.1000\text{ M}$ |
| Initial buret reading | $0.30\text{ mL}$ |
| Final buret reading | $25.30\text{ mL}$ |

Before the initial reading is recorded, an air bubble is trapped in the buret tip. During the titration, the bubble is expelled and the tip fills completely with the titrant. The student uses the difference between the final and initial buret readings to calculate the molarity of the acid. Which of the following correctly describes how the air bubble affects the calculated molarity of $\text{HA(aq)}$?

Accepted Answer

The calculated molarity is too high because the recorded volume includes titrant that fills the buret tip but does not enter the flask.

Answer

The calculated molarity is too high because expelling the air bubble increases the concentration of the $\text{NaOH(aq)}$ that subsequently enters the flask.

Answer

The calculated molarity is too low because some of the recorded titrant volume fills the buret tip, causing the student to use too few moles of $\text{NaOH}$ in the calculation.

Answer

The calculated molarity is unchanged because subtracting the initial buret reading from the final buret reading eliminates the effect of the air bubble.

Question 8

A student determines the concentration of a monoprotic acid, $\text{HA(aq)}$, by titrating a $25.00\text{ mL}$ sample with $0.1000\text{ M}$ $\text{NaOH(aq)}$. The endpoint is assumed to coincide with the equivalence point.

Before each trial, the student rinses the buret with deionized water, drains it, and then fills it directly with the $\text{NaOH(aq)}$ titrant. A small, reproducible volume of water remains in the buret and mixes with the titrant. The student calculates the acid concentration using the labeled $0.1000\text{ M}$ titrant concentration and the recorded buret volume.

How does rinsing the buret only with deionized water affect the calculated concentration of $\text{HA}$?

Accepted Answer

The calculated concentration is too high because the residual water dilutes the $\text{NaOH(aq)}$, so a larger buret volume is required to deliver the equivalence-point moles of $\text{OH}^{-}\text{(aq)}$, and that larger volume is multiplied by $0.1000\text{ M}$.

Answer

The calculated concentration is too low because the residual water lowers the concentration of the $\text{NaOH(aq)}$ delivered from the buret.

Answer

The calculated concentration is too high because the residual water dilutes the acid in the flask, so more titrant must be added to neutralize the acid.

Answer

The calculated concentration is unchanged because adding deionized water does not change the number of moles of $\text{HA}$ originally present in the flask.

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