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Chapter 9: Problem 6

The following data for the titration of a monoprotic weak acid with a strong base were collected using an automatic titrator. Prepare normal, first derivative, second derivative, and Gran plot titration curves for this data, and locate the equivalence point for each. $$ \begin{array}{cccc} \text { Volume of } \mathrm{NaOH}(\mathrm{ml}) & \mathrm{pH} & \text { Volume of } \mathrm{NaOH}(\mathrm{mL}) & \mathrm{pH} \\ \hline 0.25 & 3.0 & 49.95 & 7.8 \\ 0.86 & 3.2 & 49.97 & 8.0 \\ 1.63 & 3.4 & 49.98 & 8.2 \\ 2.72 & 3.6 & 49.99 & 8.4 \\ 4.29 & 3.8 & 50.00 & 8.7 \\ 6.54 & 4.0 & 50.01 & 9.1 \\ 9.67 & 4.2 & 50.02 & 9.4 \\ 13.79 & 4.4 & 50.04 & 9.6 \\ 18.83 & 4.6 & 50.06 & 9.8 \\ 24.47 & 4.8 & 50.10 & 10.0 \\ 30.15 & 5.0 & 50.16 & 10.2 \\ 35.33 & 5.2 & 50.25 & 10.4 \\ 39.62 & 5.4 & 50.40 & 10.6 \\ 42.91 & 5.6 & 50.63 & 10.8 \\ 45.28 & 5.8 & 51.01 & 11.0 \\ 46.91 & 6.0 & 51.61 & 11.2 \\ 48.01 & 6.2 & 52.58 & 11.4 \\ 48.72 & 6.4 & 54.15 & 11.6 \\ 49.19 & 6.6 & 56.73 & 11.8 \\ 49.48 & 6.8 & 61.11 & 12.0 \\ 49.67 & 7.0 & 68.83 & 12.2 \\ 49.79 & 7.2 & 83.54 & 12.4 \\ 49.87 & 7.4 & 116.14 & 12.6 \\ 49.92 & 7.6 & & \end{array} $$

Short Answer

Expert verified
The equivalence point occurs around 50 mL of NaOH.

Step by step solution

01

Plot Normal Titration Curve

To get started with the titration curves, plot the Volume of NaOH (mL) on the x-axis against the pH on the y-axis using the provided data set. This will give you the normal titration curve which displays how pH changes with the volume of titrant added. Identify the steepest point of this curve, which is typically near the equivalence point.
02

First Derivative Curve

Calculate the first derivative, \(\frac{dpH}{dV}\), using finite differences for the given data points. Use the differences between successive pH values divided by the differences between successive volumes of NaOH to get an estimate. Plot these values against the associated volumes of NaOH. The peak of this curve corresponds to the equivalence point.
03

Second Derivative Curve

Calculate the second derivative by taking the differences between successive first derivative values divided by the differences in the NaOH volumes. Plot these values. This curve crosses the x-axis at the equivalence point, making it easier to locate the equivalence point precisely.
04

Gran Plot

For the Gran plot, plot \(V_b \times 10^{-pH}\) versus \(V_b\) where \(V_b\) is the volume of base added. It generally applies to data just before the equivalence point where most of added \(OH^-\) ions are reacting with acid rather than changing pH predominantly. The linear portion of the plot is extrapolated to the x-axis to determine the equivalence point (volume of \(NaOH\)).
05

Identify Equivalence Points

From the normal titration curve, first and second derivative plots, pinpoint the volume of NaOH at the equivalence point as the location of maximum slope or zero crossings in respective plots. For the Gran plot, identify where the linear portion of the plot meets the x-axis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Equivalence Point
In a titration process, the equivalence point is a crucial stage. It is the point where the amount of titrant added exactly neutralizes the acid or base in the solution. This means that the number of moles of hydrogen ions equals the number of moles of hydroxide ions (for acid-base titrations). A key characteristic is that a fast change in pH will be observed around this point.

For weak acid titrations, the equivalence point typically occurs at a pH greater than 7. This is because the conjugate base of the weak acid reacts with water to form hydroxide ions, creating a slightly basic solution. Identifying the equivalence point accurately is essential, as it allows for calculations of unknown concentrations and lends insights into the acid-base properties of the solutions involved.
Derivative Plots
When performing titrations, derivative plots offer a method to pinpoint the equivalence point with higher precision. The first derivative plot involves calculating the slope, \((\frac{dpH}{dV})\), which is the rate of change of pH with respect to the volume of titrant added. This plot will typically show a peak at the equivalence point.

The second derivative plot further refines this by plotting the change in the rate of change (\

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Most popular questions from this chapter

Calculate or sketch titration curves for the following acid-base titrations. a. \(25.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NaOH}\) with \(0.0500 \mathrm{M} \mathrm{HCl}\) b. \(50.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{HCOOH}\) with \(0.100 \mathrm{M} \mathrm{NaOH}\) c. \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{NH}_{3}\) with \(0.100 \mathrm{M} \mathrm{HCl}\) d. \(50.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M}\) ethylenediamine with \(0.100 \mathrm{M} \mathrm{HCl}\) e. \(50.0 \mathrm{~mL}\) of \(0.0400 \mathrm{M}\) citric acid with \(0.120 \mathrm{M} \mathrm{NaOH}\) f. \(50.0 \mathrm{~mL}\) of \(0.0400 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) with \(0.120 \mathrm{M} \mathrm{NaOH}\)

A 0.1036 -g sample that contains only \(\mathrm{BaCl}_{2}\) and \(\mathrm{NaCl}\) is dissolved in \(50 \mathrm{~mL}\) of distilled water. Titrating with \(0.07916 \mathrm{M} \mathrm{AgNO}_{3}\) requires \(19.46 \mathrm{~mL}\) to reach the Fajans end point. Report the \(\% \mathrm{w} / \mathrm{w} \mathrm{BaCl}_{2}\) in the sample.

A quantitative analysis for aniline \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, K_{\mathrm{b}}=3.94 \times 10^{-10}\right)\) is carried out by an acid-base titration using glacial acetic acid as the solvent and \(\mathrm{HClO}_{4}\) as the titrant. A known volume of sample that contains \(3-4 \mathrm{mmol}\) of aniline is transferred to a \(250-\mathrm{mL}\) Erlenmeyer flask and diluted to approximately \(75 \mathrm{~mL}\) with glacial acetic acid. Two drops of a methyl violet indicator are added, and the solution is titrated with previously standardized \(0.1000 \mathrm{M} \mathrm{HClO}_{4}\) (prepared in glacial acetic acid using anhydrous \(\mathrm{HClO}_{4}\) ) until the end point is reached. Results are reported as parts per million aniline. (a) Explain why this titration is conducted using glacial acetic acid as the solvent instead of using water. (b) One problem with using glacial acetic acid as solvent is its relatively high coefficient of thermal expansion of \(0.11 \% /{ }^{\circ} \mathrm{C}\). For example, \(100.00 \mathrm{~mL}\) of glacial acetic acid at \(25^{\circ} \mathrm{C}\) occupies \(100.22 \mathrm{~mL}\) at \(27^{\circ} \mathrm{C}\). What is the effect on the reported concentration of aniline if the standardization of \(\mathrm{HClO}_{4}\) is conducted at a temperature that is lower than that for the analysis of the unknown? (c) The procedure calls for a sample that contains \(3-4\) mmoles of aniline. Why is this requirement necessary?

The thickness of the chromium plate on an auto fender is determined by dissolving a \(30.0-\mathrm{cm}^{2}\) section in acid and oxidizing \(\mathrm{Cr}^{3+}\) to \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) with peroxydisulfate. After removing excess peroxydisulfate by boiling, \(500.0 \mathrm{mg}\) of \(\mathrm{Fe}\left(\mathrm{NH}_{4}\right)_{2}\left(\mathrm{SO}_{4}\right)_{2} \bullet 6 \mathrm{H}_{2} \mathrm{O}\) is added, reducing the \(\mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}\) to \(\mathrm{Cr}^{3+}\). The excess \(\mathrm{Fe}^{2+}\) is back titrated, requiring \(18.29 \mathrm{~mL}\) of 0.00389 \(\mathrm{M} \mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) to reach the end point. Determine the average thickness of the chromium plate given that the density of \(\mathrm{Cr}\) is \(7.20 \mathrm{~g} / \mathrm{cm}^{3}\).

Schwartz published the following simulated data for the titration of a \(1.02 \times 10^{-4} \mathrm{M}\) solution of a monoprotic weak acid \(\left(\mathrm{p} K_{\mathrm{a}}=8.16\right)\) with \(1.004 \times 10^{-3} \mathrm{M} \mathrm{NaOH} .{ }^{10}\) The simulation assumes that a \(50-\mathrm{mL}\) pipet is used to transfer a portion of the weak acid solution to the titration vessel. A calibration of the pipet shows that it delivers a volume of only 49.94 mL. Prepare normal, first derivative, second derivative, and Gran plot titration curves for this data, and determine the equivalence point for each. How do these equivalence points compare to the expected equivalence point? Comment on the utility of each titration curve for the analysis of very dilute solutions of very weak acids. $$ \begin{array}{cccc} \mathrm{mL} \text { of } \mathrm{NaOH} & \mathrm{pH} & \mathrm{mL} \text { of } \mathrm{NaOH} & \mathrm{pH} \\ \hline 0.03 & 6.212 & 4.79 & 8.858 \\ 0.09 & 6.504 & 4.99 & 8.926 \end{array} $$ $$ \begin{array}{cccc} \mathrm{mL} \text { of } \mathrm{NaOH} & \mathrm{pH} & \mathrm{mL} \text { of } \mathrm{NaOH} & \mathrm{pH} \\ \hline 0.29 & 6.936 & 5.21 & 8.994 \\ 0.72 & 7.367 & 5.41 & 9.056 \\ 1.06 & 7.567 & 5.61 & 9.118 \\ 1.32 & 7.685 & 5.85 & 9.180 \\ 1.53 & 7.776 & 6.05 & 9.231 \\ 1.76 & 7.863 & 6.28 & 9.283 \\ 1.97 & 7.938 & 6.47 & 9.327 \\ 2.18 & 8.009 & 6.71 & 9.374 \\ 2.38 & 8.077 & 6.92 & 9.414 \\ 2.60 & 8.146 & 7.15 & 9.451 \\ 2.79 & 8.208 & 7.36 & 9.484 \\ 3.01 & 8.273 & 7.56 & 9.514 \\ 3.19 & 8.332 & 7.79 & 9.545 \\ 3.41 & 8.398 & 7.99 & 9.572 \\ 3.60 & 8.458 & 8.21 & 9.599 \\ 3.80 & 8.521 & 8.44 & 9.624 \\ 3.99 & 8.584 & 8.64 & 9.645 \\ 4.18 & 8.650 & 8.84 & 9.666 \\ 4.40 & 8.720 & 9.07 & 9.688 \\ 4.57 & 8.784 & 9.27 & 9.706 \end{array} $$
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