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Chapter 7: Problem 83

A \(40.0 \mathrm{~mL}\) solution of weak base, \(\mathrm{BOH}\) is titrated with \(0.1 \mathrm{~N} \mathrm{HCl}\) solution. The \(\mathrm{pH}\) of the solution is found to be \(10.04\) and \(9.14\) after adding \(5.0 \mathrm{~mL}\) and \(20.0 \mathrm{~mL}\) of the acid respectively. Find out the dissociation constant of the base.

Short Answer

Expert verified
The dissociation constant \( K_b \) can be calculated by solving for \( \text{p}K_b \) using the provided pH values and the Henderson-Hasselbalch equation.

Step by step solution

01

Write down the Neutralization Reaction

The reaction between the weak base \( \text{BOH} \) and the strong acid \( \text{HCl} \) is: \[ \text{BOH} + \text{HCl} \rightarrow \text{BCl} + \text{H}_2\text{O} \]. This reaction causes the acidic and basic properties to neutralize each other.
02

Calculate Moles of HCl Added

The normality (N) of the HCl solution is given as 0.1 N. The moles of HCl added can be determined using the formula \( \text{moles} = \text{Normality} \times \text{Volume (L)} \). For the first point, \( \text{moles} = 0.1 \times 0.005 = 0.0005 \) moles.
03

Henderson-Hasselbalch Equation for First Point

In the case of weak bases, the Henderson-Hasselbalch equation is given by \( \text{pH} = \text{p}K_b + \log\left(\frac{[\text{BOH}]}{[\text{BOH}_2^+]}\right) \). After adding 5 mL of HCl, \( \text{pH} = 10.04 \). Use this to set up the equation: \( 10.04 = \text{p}K_b + \log\left(\frac{[\text{BOH}]}{0.0005}\right) \).
04

Calculate Moles of HCl at Second Point

After adding 20 mL of HCl, the moles of HCl are \( 0.1 \times 0.020 = 0.002 \) moles.
05

Henderson-Hasselbalch Equation for Second Point

At this stage, \( \text{pH} = 9.14 \). Use the Henderson-Hasselbalch equation again: \( 9.14 = \text{p}K_b + \log\left(\frac{[\text{BOH}]}{0.002}\right) \).
06

Solve System of Equations

Using the equations from Steps 3 and 5, subtract them to find \( \frac{[\text{BOH}]}{[\text{BOH}_2^+]} \). This will eliminate \( \text{p}K_b \). Solve the equation to find \( \text{p}K_b \).
07

Calculate the Dissociation Constant \( K_b \)

Once \( \text{p}K_b \) is found, use the relationship \( \text{p}K_b = -\log( K_b ) \). Solve for \( K_b \) by taking the antilog: \( K_b = 10^{-\text{p}K_b} \).

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Key Concepts

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

Henderson-Hasselbalch Equation
The Henderson-Hasselbalch equation offers a simplified way to estimate the pH of a solution during a titration, especially when dealing with weak bases. For weak bases, the equation is slightly modified from its more commonly known form for acids. The equation reads:\[ \text{pH} = \text{p}K_b + \log\left(\frac{[\text{Base}]}{[\text{Conjugate Acid]}]}\right) \]In this context, the base is \( \text{BOH} \) and the conjugate acid is \( \text{BOH}_2^+ \). This equation helps us relate the ratio of the concentrations of the base and its conjugate acid to the pH measurements observed during the experiment.
This is important because it allows the calculation of the base's dissociation constant through the known values of pH and the volumes of acid added.
Neutralization Reaction
A neutralization reaction between a weak base and a strong acid results in the formation of water and a salt. In this exercise, the weak base \(\text{BOH}\) reacts with the strong acid \(\text{HCl}\).
The reaction is represented as:\[ \text{BOH} + \text{HCl} \rightarrow \text{BCl} + \text{H}_2\text{O} \]Through such a reaction, acids and bases neutralize each other. The role of HCl here is crucial as it donates protons (\(\text{H}^+\)), which the weak base can accept to form its conjugate acid \(\text{BOH}_2^+\).
  • This highlights the transformation of the base due to the addition of the acid.
  • It shows how the titration changes the chemical equilibrium.
Understanding this reaction aids in comprehending the changes in the concentration of reactive species during the process.
Weak Bases
A weak base is one that does not fully dissociate in solution. In contrast to strong bases, weak bases only partially ionize, meaning a portion of the base remains as the original molecule.
For \(\text{BOH}\), the equilibrium in water can be expressed as:\[ \text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^- \]Challenges with weak bases include calculating the equilibrium concentrations and understanding their interaction with added acids.
  • They result in a buffer with their conjugate acids.
  • The degree of ionization is normally represented by the dissociation constant \(K_b\).
Such bases are particularly useful in buffer solutions, where their capacity to neutralize added acids (without significantly changing the pH) makes them valuable.
pH Calculation
Calculating the pH is fundamental in understanding acid-base reactions and their consequences. Specifically, for weak bases, you measure the pH to obtain insight into the base's strength and the progression of the titration.The pH of a solution is determined by the concentration of hydrogen ions (\(\text{H}^+\)). During the above experiment:
  • The Henderson-Hasselbalch equation was applied to find the relationship between pH and \(K_b\), allowing the calculation of the dissociation constant of \(\text{BOH}\).
  • Initial and subsequent pH values \(10.04\ and \9.14\) provided the necessary empirical data.
The pH scale is logarithmic, meaning each whole number change equates to a tenfold change in ion concentration.
For this reason, small changes in pH can reflect significant changes in concentrations of reactive species.

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Most popular questions from this chapter

An acidic buffer solution can be prepared by mixing the solutions of (a) ammonium acetate and acetic acid (b) ammonium chloride and ammonioum hydroxide (c) sulphuric acid and sodium sulphate (d) sodium chloride and sodium hydroxide.

$$ \begin{aligned} &\mathrm{NaOH} \text { is a strong base. What will be } \mathrm{pH} \text { of } 5.0 \times 10^{-2} \mathrm{M} \mathrm{NaOH}\\\ &\text { solution ? }(\log 2=0.3) \end{aligned} $$ (a) \(14.00\) (b) \(13.70\) (c) \(13.00\) (d) \(12.70\)

The Haber's process for the formation of \(\mathrm{NH}_{3}\) at \(298 \mathrm{~K}\) is \(\mathrm{N}_{2}+3 \mathrm{H}_{2} \rightleftharpoons 2 \mathrm{NH}_{3} ; \Delta H=-46.0 \mathrm{~kJ} ;\) Which of the following is the correct statement (a) The condition for equilibrium is $$ G_{\mathrm{N}_{2}}+3 G_{\mathrm{H}_{2}}=2 G_{\mathrm{NH}_{3}} $$ where \(G\) is Gibb's free energy per mole of the gaseous species measured at that partial pressure. (b) On adding \(\mathrm{N}_{2}\), the equilibrium will shift to forward direction because according to \(\mathrm{II}^{\text {nd }}\) law of thermodynamics, the entropy must increase in the direction of spontaneous reaction (c) The catalyst will increase the rate of forward reaction by 2 times and that of backward reaction by \(1.5\) times (d) None of these

For the following Assertion and Reason, the correct option is: Assertion: The \(\mathrm{pH}\) of water increases with increase in temperature. Reason: The dissociation of water into \(\mathrm{H}^{+}\)and \(\mathrm{OH}^{-}\)is an exothermic reaction. (a) Both assertion and reason are true, and the reason is the correct explanation for the assertion. (b) Both assertion and reason are false. (c) Both assertion and reason are true, but the reason is not the correct explanation for the assertion. (d) Assertion is not true, but reason is true.

Consider the following reaction: \(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g}) ; \mathrm{DH}^{0}=+58 \mathrm{~kJ}\) For each of the following cases ((i), (ii)), the direction in which the equilibrium shifts is : (i) Temperature is decreases (ii) Pressure is increased by adding \(\mathrm{N}_{2}\) at constant \(\mathrm{T}\). (a) (i) towards product, (ii) towards product (b) (i) towards reactant, (ii) towards product (c) (i) towards reactant, (ii) no change (d) (i) towards product, (ii) no change
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