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Chapter 5: Problem 31

The pKa pf HCN is \(9.30 .\) The pH of a solution prepared by mixing \(2.5\) moles of \(\mathrm{KCN}\) and \(2.5\) moles of HCN in water and making up the total volume to \(500 \mathrm{ml}\) is a. \(9.30\) b. \(8.30\) c. \(7.30\) d. \(10.30\)

Short Answer

Expert verified
The pH of the solution is 9.30.

Step by step solution

01

Understand the Components

We have a mixture of \(2.5\) moles of \(\mathrm{KCN}\) and \(2.5\) moles of \(\mathrm{HCN}\), forming a buffer solution. \(\mathrm{KCN}\) provides \(\mathrm{CN}^-\), the conjugate base of \(\mathrm{HCN}\). This means we have equal concentrations of the weak acid \(\mathrm{HCN}\) and its conjugate base \(\mathrm{CN}^-\).
02

Use the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation for a buffer solution is given by: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]}\right ) \]Since \([\text{A}^-] = [\text{HA}]\) in this case (because both are \(2.5/0.5\) mol/L when considering the 500 mL solution), the log term becomes \(\log(1) = 0\).
03

Calculate the pH

Substitute the given \(\text{pKa} = 9.30\) and the calculated value from the ratio into the Henderson-Hasselbalch equation:\[ \text{pH} = 9.30 + \log(1) = 9.30 + 0 = 9.30 \].

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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 is a crucial tool in chemistry for calculating the pH of buffer solutions. A buffer solution is composed of a weak acid and its conjugate base or a weak base and its conjugate acid.
This equation is especially valuable because it allows you to predict the pH of a solution without the need for extensive calculations or experimental data.
  • It is represented mathematically as: \[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
  • The symbol \(\text{pKa}\) denotes the acid dissociation constant, a measure of the strength of the acid.
  • \([\text{A}^-]\) is the concentration of the conjugate base, while \([\text{HA}]\) is the concentration of the un-dissociated weak acid.
The equation assumes that changes in the concentrations of the acid and base within the buffer are relatively small, which simplifies calculations.
By using the Henderson-Hasselbalch equation, you can quickly determine how a buffer will respond to the addition of an acid or base.
Conjugate Acid-Base Pair
In buffer solutions, the concept of conjugate acid-base pairs is central. A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton.
This relationship helps maintain the buffer鈥檚 pH within a narrow range.
  • The acid member of the pair can donate a proton, turning into its conjugate base. Similarly, the base member can accept a proton, turning into its conjugate acid.
  • For example, in the exercise, \(\text{HCN}\) (the weak acid) and \(\text{CN}^-\) (the conjugate base) form a conjugate acid-base pair.
  • When added to water, the \(\text{HCN}\) partially dissociates to produce \(\text{H}^+\) ions and \(\text{CN}^-\) ions.
This ability to donate and accept protons allows buffers to mitigate changes in pH when acids or bases are added.
Conjugate pairs are a backbone in understanding how buffers function effectively.
pH Calculation
Calculating the pH of a solution is a fundamental task in chemistry, especially when dealing with buffer solutions. The pH is a measure of the hydrogen ion concentration in a solution and dictates the acidity or basicity of that solution.
In the context of buffer solutions, where an acid and its conjugate base exist in equilibrium, using straightforward calculations can provide immediate insight into the system's properties.
  • The pH scale ranges from 0 to 14, with 7 being neutral. Values below 7 indicate acidic conditions, while values above 7 mark basic conditions.
  • In our example, the buffer solution formed by mixing equal moles of \(\text{HCN}\) and \(\text{CN}^-\) results in a pH that approximately equals the pKa of the weak acid. This is a typical property of equimolar buffer solutions.
  • The order of calculations involves simply substituting into the Henderson-Hasselbalch equation: \[ \text{pH} = \text{pKa} + \log \left( \frac{2.5}{2.5} \right) = \text{pKa} + \log(1) = \text{pKa}. \]
This straightforward calculation reveals that the buffer solution is able to resist changes in pH effectively, an essential property particularly useful in various scientific applications.

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Most popular questions from this chapter

The \(\mathrm{pH} 0.1 \mathrm{M}\) solution the following salts increases in the order a. \(\mathrm{NaCl}<\mathrm{NH}_{4} \mathrm{Cl}<\mathrm{NaCN}<\mathrm{HCl}\) b. \(\mathrm{HCl}<\mathrm{NH}_{4} \mathrm{Cl}<\mathrm{NaCl}<\mathrm{NaCN}\) c. \(\mathrm{NaCN}<\mathrm{NH}_{4} \mathrm{Cl}<\mathrm{NaCl}<\mathrm{HCl}\) d. \(\mathrm{HCl}<\mathrm{NaCl}<\mathrm{NaCN}<\mathrm{NH}_{4} \mathrm{Cl}\)

\((\mathrm{A}): \mathrm{H}_{3} \mathrm{PO}_{3}\) is a dibasic acid. \((\mathbf{R})\) : There are two hydrogen atoms directly attached to phosphorus

Which of the following statement is/are incorrect here? a. Equilibrium constant of an exothermic reaction decreases with the increase of temperature b. \(\mathrm{K}_{\mathrm{p}}\) is always greater than \(\mathrm{K}_{\mathrm{C}}\)

What is the correct sequence of active masses in increasing order in gaseous mixture containing one gram per litre of each of the following? I. \(\mathrm{NH}_{3}\) II. \(\mathrm{N}_{2}\) III. \(\mathrm{H}_{2}\) IV. \(\mathrm{O}_{2}\) Select the correct answer using the codes given below: a. III, I, IV, II b. III, IV, II, I c. II, I, IV, III d. IV, II, I, III

The solubility of \(\mathrm{A}_{2} \mathrm{X}_{3}\) is \(\mathrm{y} \mathrm{mol} \mathrm{dm}^{-3} .\) Its solubility product is a. \(6 y^{4}\) b. \(64 \mathrm{y}^{4}\) c. \(36 \mathrm{y}^{5}\) d. \(108 \mathrm{y}^{5}\)
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