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Chapter 15: Problem 64

Consider the titration of 80.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) by 0.400 \(\mathrm{M} \mathrm{HCl}\) . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of HCl have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}} \\ {\text { c. } 30.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}}\end{array} $$

Short Answer

Expert verified
The pH of the resulting solution after the addition of different volumes of HCl is as follows: a. \(0.0 \, \mathrm{mL}: \) pH = 13.70 b. \(20.0 \, \mathrm{mL}: \) pH = 7 c. \(30.0 \, \mathrm{mL}: \) pH = 1.30 d. \(40.0 \, \mathrm{mL}: \) pH = 0.70 e. \(80.0 \, \mathrm{mL}: \) pH = 0.10

Step by step solution

01

Calculate the initial moles of Ba(OH)鈧�

First, we need to calculate the initial moles of Barium Hydroxide (Ba(OH)鈧�) in the solution using the given concentration and volume: Moles of Ba(OH)鈧� = Molarity 脳 Volume Moles of Ba(OH)鈧� = 0.100 mol/L 脳 0.080 L = 0.008 mol
02

Calculate the moles of HCl added at each point

For each volume of HCl added, we will calculate the moles of HCl using its concentration and the given volume. We will do that with the following formula: Moles of HCl = Molarity 脳 Volume For example, for volume a (0.0 mL), we have: Moles of HCl = 0.400 mol/L 脳 0 mL = 0 mol Repeat this for each additional volume of HCl provided.
03

Create and fill in the table

Next, we will create and fill in a table to compare the moles of Ba(OH)鈧� and HCl at different stages of the titration. | Volume of HCl added | Moles of Ba(OH)鈧� | Moles of HCl added | Moles of reacted Ba(OH)鈧� | Moles of unreacted Ba(OH)鈧� | Moles of remaining HCl | |---------------------|------------------|--------------------|-------------------------|---------------------------|-----------------------| | a. 0.0 mL | 0.008 | 0 | 0 | 0.008 | 0 | | b. 20.0 mL | 0.008 | 0.008 | 0.008 | 0 | 0 | | c. 30.0 mL | 0.008 | 0.012 | 0.008 | 0 | 0.004 | | d. 40.0 mL | 0.008 | 0.016 | 0.008 | 0 | 0.008 | | e. 80.0 mL | 0.008 | 0.032 | 0.008 | 0 | 0.024 | We will now explain in detail how to fill in each row of the table. - 1. Write the moles of Ba(OH)鈧� at the start - 2. Write the moles of HCl added in each scenario - 3. Calculate the moles of reacted Ba(OH)鈧� by taking the minimum of the moles of Ba(OH)鈧� and moles of HCl added - 4. Calculate the moles of unreacted Ba(OH)鈧� by subtracting the moles of reacted Ba(OH)鈧� from the initial moles of Ba(OH)鈧� (if any) - 5. Calculate the moles of remaining HCl by subtracting the moles of reacted Ba(OH)鈧� from the moles of HCl added (if any)
04

Determine the pH of the solution

Lastly, based on the results in the table, we will calculate the pH of the solution at each stage of the titration. It is important to analyze in which phase the solution is, whether it is before the equivalence point, at the equivalence point, or after the equivalence point. This can be established through the comparison of moles of unreacted Ba(OH)鈧� and remaining HCl in the table: a. pH = 14 - pOH Since we have unreacted Ba(OH)鈧�, pOH=-log[OH鈦籡, calculate the concentration of OH鈦� ions from unreacted Ba(OH)鈧� and find the pH. b. At equivalence point, the pH will be neutral (pH = 7) as the strong acid fully reacts with the strong base. c,d,e. Since there is excess HCl in the solution, directly calculate the pH using the H鈦� ion concentration from the remaining moles of HCl: pH = -log[H鈦篯 By following these steps, we can analyze the titration and find the pH at different stages of the process.

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Key Concepts

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

pH Calculation
The pH calculation is crucial in understanding the acidity or basicity of a solution during a titration. pH is the measure of hydrogen ion concentration in a solution, represented as:\[pH = -\log[H^+]\]For a titration, it helps to determine the point of neutralization between the acid and base. In the case of a solution containing a strong base like Ba(OH)鈧� and a strong acid like HCl, calculations differ based on the phase of the titration:
  • Before the equivalence point, the solution contains unreacted base, and the pH can be found using the concentration of OH鈦� ions.
  • At the equivalence point, the pH is usually around 7, as the moles of acid and base are equivalent.
  • After the equivalence point, excess acid determines the pH, calculating directly from the H鈦� concentration.
Understanding how to calculate pH at each stage allows for a comprehensive analysis of the titration process.
Acid-Base Reaction
An acid-base reaction, such as the titration of Ba(OH)鈧� with HCl, involves the transfer of protons (H鈦� ions) between the acid and the base. The reaction is given by:\[\text{Ba(OH)}_2 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O}\]Here, the strong base Barium Hydroxide neutralizes the strong acid Hydrochloric acid. Key elements to understand in this reaction are:
  • The stoichiometry indicates the proportions required for complete neutralization, i.e., two moles of HCl react with one mole of Ba(OH)鈧�.
  • Products formed are water (H鈧侽) and a salt (BaCl鈧�), which importantly, influence the pH level post-reaction.
  • Following the entire titration process helps in predicting behavior during partial or excess neutralization stages.
This reaction represents a basic model of how acids and bases interact during titration experiments.
Equivalence Point
The equivalence point is a key milestone in titration, where the amount of acid equals the amount of base in a solution. This means:
  • All moles of the base have reacted completely with the moles of acid.
  • In our specific scenario, it occurs when exactly 2 moles of HCl have reacted for each mole of Ba(OH)鈧�.
  • At this point, the solution is typically neutral, resulting in a pH around 7, although this can vary depending on the strength of acid and base.
Detecting the equivalence point can be done through indicators, pH meters, or by observing the point where an abrupt change in pH occurs during the titration.
Molarity
Molarity is fundamental in determining concentration aspects of a solution, expressed as the number of moles of solute per liter of solution (mol/L). It guides the calculation of moles of reactants necessary for titration. Suppose you have:
  • 0.100 M Ba(OH)鈧� indicates 0.1 moles of Barium Hydroxide in one liter of solution.
  • 0.400 M HCl similarly represents 0.4 moles of Hydrochloric Acid per liter.
Understanding molarity allows you to:
  • Determine how much of each reactant is in the solution before any reaction occurs.
  • Translate volumes of titrant added into moles for stoichiometry checks during titration.
This concept is central to precisely conducting and analyzing titration results.
Stoichiometry
Stoichiometry involves the calculation of reactants and products in chemical reactions based on balanced chemical equations. For the titration of Ba(OH)鈧� with HCl, the reaction can be summarized by:\[\text{Ba(OH)}_2 + 2 \text{HCl} \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O}\]Points to consider include:
  • The stoichiometric coefficients indicate that one mole of Ba(OH)鈧� reacts with two moles of HCl.
  • This ratio is vital for determining the progression through titration phases - before, at, and after the equivalence point.
  • Using stoichiometry, you can predict moles of Ba(OH)鈧� and HCl remaining, guiding pH calculations.
Indeed, stoichiometry forms the backbone of any titration process, ensuring all calculations align with theoretical predictions.

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Most popular questions from this chapter

A \(0.400-M\) solution of ammonia was titrated with hydrochloric acid to the equivalence point, where the total volume was 1.50 times the original volume. At what pH does the equivalence point occur?

Calculate the pH after 0.15 mole of solid NaOH is added to 1.00 \(\mathrm{L}\) of each of the following solutions: a. 0.050\(M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{5} \mathrm{O}_{2}, K_{2}=1.3 \times 10^{-5}\right)\) and 0.080\(M\) sodium propanoate b. 0.50\(M\) propanoic acid and 0.80\(M\) sodium propanoate c. Is the solution in part a still a buffer solution after the NaOH has been added? Explain.

Calculate the pH of a solution formed by mixing 100.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M}\) NaF and 100.0 \(\mathrm{mL}\) of 0.025 \(\mathrm{M} \mathrm{HCl} .\)

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) \(\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)\) by 0.200\(M \mathrm{HNO}_{3}\) . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 50.0 \mathrm{mL}} \\ {\text { c. } 25.0 \mathrm{mL}} & {\text { f. } 100.0 \mathrm{mL}}\end{array} $$

Tris (hydroxymethyl)aminomethane, commonly called TRIS or Trizma, is often used as a buffer in biochemical studies. Its buffering range is \(\mathrm{pH} 7\) to \(9,\) and \(K_{\mathrm{b}}\) is \(1.19 \times 10^{-6}\) for the aqueous reaction $$ \left(\mathrm{HOCH}_{2}\right)_{3} \mathrm{CNH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons\left(\mathrm{HOCH}_{2}\right)_{3} \mathrm{CNH}_{3}^{+}+\mathrm{OH}^{-} $$ a. What is the optimal pH for TRIS buffers? b. Calculate the ratio \([T R I S] /\left[T R I S H^{+}\right]\) at \(p H=7.00\) and at p H=9.00 c. A buffer is prepared by diluting 50.0 \(\mathrm{g}\) TRIS base and 65.0 \(\mathrm{g}\) TRIS hydrochloride (written as TRISHCl) to a total volume of 2.0 \(\mathrm{L}\) What is the pH of this buffer? What is the pH after 0.50 \(\mathrm{mL}\) of 12 \(\mathrm{MHCl}\) is added to a 200.0 -mL portion of the buffer?
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