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Chapter 16: Problem 21

Calculate the \(\mathrm{pH}\) during the titration of \(1.00 \mathrm{~mL}\) of \(0.240 \mathrm{M} \mathrm{LiOH}\) with \(0.200 \mathrm{M} \mathrm{HNO}_{3}\) after 0,0.25,0.50 \(1.20,\) and \(1.50 \mathrm{~mL}\) nitric acid have been added. Sketch the titration curve.

Short Answer

Expert verified
The pH changes from 13.38 to 1.62 as HNO鈧� is added.

Step by step solution

01

Understanding the Reaction

The titration involves the reaction between lithium hydroxide (LiOH), a strong base, and nitric acid (HNO鈧�), a strong acid. The reaction is: \[ \text{LiOH} + \text{HNO}_3 \rightarrow \text{LiNO}_3 + \text{H}_2\text{O} \]The neutralization reaction results in the production of lithium nitrate and water.
02

Calculate Initial pH (0 mL HNO鈧� added)

Initially, there is only LiOH in the solution. The concentration of OH鈦� ions is the same as the LiOH concentration because it is a strong base.\[ \text{[OH}^-\text{]} = 0.240 \text{ M} \]To find the pH, first calculate the pOH:\[ \text{pOH} = -\log(0.240) \approx 0.62 \]Then use the relation: \[ \text{pH} = 14.00 - \text{pOH} = 14.00 - 0.62 = 13.38 \]
03

Calculate pH after 0.25 mL HNO鈧� added

Calculate the moles of HNO鈧� added:\[ 0.25 \text{ mL} \times \frac{0.200 \text{ mol}}{1000 \text{ mL}} = 0.00005 \text{ mol} \]Initial moles of LiOH:\[ 1.00 \text{ mL} \times \frac{0.240 \text{ mol}}{1000 \text{ mL}} = 0.00024 \text{ mol} \]Moles of OH鈦� remaining after neutralization:\[ 0.00024 - 0.00005 = 0.00019 \text{ mol} \]Remaining concentration in total volume (1.25 mL):\[ \text{[OH}^-\text{]} = \frac{0.00019}{0.00125} = 0.152 \text{ M} \]Calculate pOH and then pH:\[ \text{pOH} = -\log(0.152) \approx 0.82 \]\[ \text{pH} = 14.00 - 0.82 = 13.18 \]
04

Calculate pH after 0.50 mL HNO鈧� added

Moles of HNO鈧� added:\[ 0.50 \text{ mL} \times \frac{0.200 \text{ mol}}{1000 \text{ mL}} = 0.00010 \text{ mol} \]Moles of OH鈦� remaining:\[ 0.00024 - 0.00010 = 0.00014 \text{ mol} \]Remaining concentration in total volume (1.50 mL):\[ \text{[OH}^-\text{]} = \frac{0.00014}{0.00150} = 0.0933 \text{ M} \]Calculate pOH and then pH:\[ \text{pOH} = -\log(0.0933) \approx 1.03 \]\[ \text{pH} = 14.00 - 1.03 = 12.97 \]
05

Calculate pH after 1.20 mL HNO鈧� added

Moles of HNO鈧� added:\[ 1.20 \text{ mL} \times \frac{0.200 \text{ mol}}{1000 \text{ mL}} = 0.00024 \text{ mol} \]At this point, moles of HNO鈧� = moles of LiOH, neutralizing completely.At equivalence point, the solution is neutral:\[ \text{pH} = 7.00 \]
06

Calculate pH after 1.50 mL HNO鈧� added

Moles of HNO鈧� added:\[ 1.50 \text{ mL} \times \frac{0.200 \text{ mol}}{1000 \text{ mL}} = 0.00030 \text{ mol} \]Excess moles of HNO鈧�:\[ 0.00030 - 0.00024 = 0.00006 \text{ mol} \]Concentration of excess H鈦� in total volume (2.50 mL):\[ \text{[H}^+\text{]} = \frac{0.00006}{0.00250} = 0.024 \text{ M} \]Calculate pH:\[ \text{pH} = -\log(0.024) \approx 1.62 \]
07

Sketch the Titration Curve

Create a graph plotting the volume of HNO鈧� added (x-axis) vs pH (y-axis): - Start high at pH 13.38 at 0 mL. - Gradually decrease to near-neutral at 1.20 mL (pH = 7.00). - Then sharply drop to pH = 1.62 at 1.50 mL. This is characteristic of titrations between a strong acid and a strong base.

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Key Concepts

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

pH Calculation
Calculating the pH during a titration process involves determining the concentration of hydrogen ions (H鈦�) or hydroxide ions (OH鈦�) at various points of the titration. It involves a few key steps:
  • Identify the initial concentration of the base or acid before any titrant is added. For a strong base like LiOH, this is straightforward as it completely dissociates in water.
  • Determine the moles of titrant (HNO鈧� in this case) added at each stage by using the formula: \[ ext{moles} = ext{volume (L) of titrant} \times \text{molarity of titrant} \]
  • Calculate the remaining concentration of the base (or acid) after neutralization of the added acid (or base). Since this particular example involves strong acids and bases, the calculations are direct and assume full dissociation.
  • Apply the formula for pH conversion: For remaining OH鈦� ions, first calculate pOH and then convert to pH with: \[ \text{pH} = 14.00 - \text{pOH} \]
  • For excess H鈦� ions, use: \[ \text{pH} = -\log(\text{[H}^+\text{]}) \]
By understanding these steps, you can compute the pH at any stage of a titration, helping to plot an accurate titration curve.
Titration Curve
A titration curve is a graphical representation that shows how the pH of a solution changes as a titrant is added. The titration of a strong acid with a strong base demonstrates characteristic features like a steep drop in pH. Here鈥檚 how a typical curve is constructed and interpreted:
  • Initial Region: Starts at a higher pH when a strong base is present with no titrant added. For example, in this exercise, the pH starts at 13.38 due to the strong base LiOH.

  • Buffer Region: As the acid is added, the pH decreases slowly while the base is neutralized until you approach the equivalence point.

  • Equivalence Point: A sudden drop occurs when equal amounts of acid and base have reacted, leading to a near-neutral pH of 7.00. This point is crucial as it indicates when neutralization is complete.

  • Post-Equivalence: Further addition of acid results in an excess of H鈦� ions, causing the pH to drop sharply, as seen from 1.20 mL to 1.50 mL addition, reaching a very acidic pH of 1.62.
This titration curve helps visualize pH changes and understand the chemical process dynamics between the acid and base.
Equivalence Point
In acid-base titrations, the equivalence point is a crucial concept representing the stage where the amount of acid completely neutralizes the base (or vice versa), resulting in a chemical equilibrium. For strong acid-strong base titrations like HNO鈧� and LiOH, the equivalence point has some unique characteristics:
  • It occurs when the moles of added titrant equal the moles of the substance initially present in the solution. In our example, this occurs after the addition of 1.20 mL of HNO鈧�.

  • At this point, the resulting solution is neutral with a pH typically around 7.00, because both the acid and base have been fully reacted, producing only water and a salt (lithium nitrate in this case) in the solution.

  • The dramatic change in pH near the equivalence point is a hallmark of such titrations and is often used to determine the precise point of neutralization. This is useful in calculating the unknown concentration of one of the reactants in analytical chemistry.
The equivalence point is foundational in understanding the completion of a chemical reaction during titration.

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Most popular questions from this chapter

Calculate how much \(0.100 \mathrm{M} \mathrm{NaOH}\) is needed to react completely (a) \(45.00 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{HCl}\). (b) \(5.00 \mathrm{~mL}\) of \(0.350 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) (forming \(\left.\mathrm{Na}_{2} \mathrm{SO}_{4}\right)\). (c) \(10.00 \mathrm{~mL}\) of \(0.100 \mathrm{M}\) acetic acid.

Calculate the \(\mathrm{pH}\) of each of the following solutions. (a) \(10.0 \mathrm{~mL}\) of \(0.300 \mathrm{M}\) hydrofluoric acid plus \(30.0 \mathrm{~mL}\) of \(0.100 M\) sodium hydroxide (b) \(100.0 \mathrm{~mL}\) of \(0.250 \mathrm{M}\) ammonia plus \(50.0 \mathrm{~mL}\) of \(0.100 M\) hydrochloric acid (c) \(25.0 \mathrm{~mL}\) of \(0.200 \mathrm{M}\) sulfuric acid plus \(50.0 \mathrm{~mL}\) of \(0.400 \mathrm{M}\) sodium hydroxide

How many millimoles of \(\mathrm{KOH}\) are needed to neutralize completely \(35.1 \mathrm{~mL}\) of \(0.101 \mathrm{M}\) nitric acid?

Sketch the curve for the titration of \(100 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) weak acid \(\left(K_{a}=1.0 \times 10^{-4}\right)\) with a \(0.20 M\) strong base. On the same axes, sketch the titration curve for the same volume and concentration of \(\mathrm{HCl}\).

Phenolphthalein is a commonly used indicator that is colorless in the acidic form \((\mathrm{pH}<8.3)\) and pink in the base form \((\mathrm{pH}>10.0)\). It is a weak acid with a \(\mathrm{pK}_{\mathrm{a}}\) of 8.7 . What fraction is in the acid form when the acid color is apparent? What fraction is in the base form when the base color is apparent?
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