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What are the major species present in \(0.250 M\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) b. \(\mathrm{HCN}\)

Short Answer

Expert verified
In the 0.250 M solution of phenol (HOC鈧咹鈧), the major species present are HOC鈧咹鈧, OC鈧咹鈧呪伝, and H鈧僌鈦. The pH of this solution is approximately 5.3. In the 0.250 M solution of hydrogen cyanide (HCN), the major species present are HCN, CN鈦, and H鈧僌鈦. The pH of this solution is approximately 4.96.

Step by step solution

01

Identifying the major species present in the solution of HOC鈧咹鈧

In a 0.250 M solution of HOC鈧咹鈧 (phenol), the major species present are the phenol (HOC鈧咹鈧) itself, the phenoxide ion (OC鈧咹鈧呪伝), and the hydronium ion (H鈧僌鈦).
02

Calculating the concentration of H鈧僌鈦 ions in the solution of HOC鈧咹鈧

To calculate the concentration of H鈧僌鈦 ions, we first need to determine the dissociation constant (K鈧) of phenol. Phenol has a K鈧 value of \(1.0 脳 10^{-10}\) at 25掳C. Using the formula for the ionization of weak acids: K鈧 = \(\frac{[H鈧僌鈦篯 [OC鈧咹鈧呪伝]}{[HOC鈧咹鈧匽}\) Let x represent the concentration of H鈧僌鈦 and OC鈧咹鈧呪伝 since they are equal. Then, K鈧 = \(\frac{x^2}{0.250-x}\) Solve for x (ignoring the minus x in the denominator due to the small K鈧 value): x虏 = K鈧 脳 0.250 x虏 = \(1 脳 10^{-10} 脳 0.250\) x = \(\sqrt{2.5 脳 10^{-11}}\) x = \(5 脳 10^{-6}\) M So, [H鈧僌鈦篯 = \(5 脳 10^{-6}\) M in the phenol solution.
03

Calculating the pH of the HOC鈧咹鈧 solution

Now that we have the concentration of H鈧僌鈦 ions, we can calculate the pH using the pH formula: pH = -log鈧佲個 [H鈧僌鈦篯 pH = -log鈧佲個 (\(5 脳 10^{-6}\)) pH 鈮 5.3 Now, let's do the same for HCN. b. HCN
04

Identifying the major species present in the solution of HCN

In a 0.250 M solution of HCN, the major species present are the hydrogen cyanide (HCN) itself, cyanide ion (CN鈦), and the hydronium ion (H鈧僌鈦).
05

Calculating the concentration of H鈧僌鈦 ions in the solution of HCN

We need to determine the dissociation constant (K鈧) of HCN. Hydrogen cyanide has a K鈧 value of \(4.9 脳 10^{-10}\) at 25掳C. Using the formula for the ionization of weak acids: K鈧 = \(\frac{[H鈧僌鈦篯 [CN鈦籡}{[HCN]}\) Let x represent the concentration of H鈧僌鈦 and CN鈦 since they are equal. Then, K鈧 = \(\frac{x^2}{0.250-x}\) Solve for x (ignoring the minus x in the denominator due to the small K鈧 value): x虏 = K鈧 脳 0.250 x虏 = \(4.9 脳 10^{-10} 脳 0.250\) x = \(\sqrt{1.225 脳 10^{-10}}\) x = \(1.1 脳 10^{-5}\) M So, [H鈧僌鈦篯 = \(1.1 脳 10^{-5}\) M in the HCN solution.
06

Calculating the pH of the HCN solution

Now that we have the concentration of H鈧僌鈦 ions, we can calculate the pH using the pH formula: pH = -log鈧佲個 [H鈧僌鈦篯 pH = -log鈧佲個 (\(1.1 脳 10^{-5}\)) pH 鈮 4.96 So, the major species present in the phenol solution are HOC鈧咹鈧, OC鈧咹鈧呪伝, and H鈧僌鈦 and the pH of this solution is approximately 5.30. The major species present in the HCN solution are HCN, CN鈦, and H鈧僌鈦 and the pH of this solution is approximately 4.96.

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Key Concepts

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

pH Calculation
The pH of a solution reflects its acidity or alkalinity. It is a logarithmic measure of the concentration of hydronium ions (\(H_3O^+\)) in a solution. Specifically, pH is calculated using the formula: \[ pH = -\log_{10} [H_3O^+] \] This means as the concentration of hydronium ions in the solution increases, the pH value decreases, indicating a more acidic solution. When dealing with weak acids like phenol (\(HOC_6H_5\)) and hydrogen cyanide (\(HCN\)), which don't fully dissociate in water, the pH calculation involves determining the concentration of these ions from an equilibrium expression before applying the above formula.

For educational purposes, we simplify the process by assuming that the change in the initial concentration of the acid due to ionization is negligible. Thus, the pH calculation for weak acids typically involves the square root of the product of the acid's ionization constant (\(K_a\)) and its initial concentration. This approximation holds well for weak acids with small dissociation constants and allows students to understand the concept without getting bogged down in more complex equilibrium calculations.
Weak Acid Dissociation
Weak acids do not fully dissociate in water. Instead, they establish an equilibrium between the undissociated acid and the products of its dissociation - the conjugate base and the hydronium ion. The dissociation of a generic weak acid, HA, can be represented as: \[ HA + H_2O \rightleftharpoons A^- + H_3O^+ \]

Where HA is the weak acid, A- is its conjugate base, and H3O+ represents the hydronium ion. The extent to which a weak acid dissociates is governed by its acid dissociation constant, Ka, which quantitatively describes the equilibrium concentrations of the reactants and products. For example, with phenol, only a small amount dissociates into phenoxide ions (\(OC_6H_5^-\)) and hydronium ions, which makes phenol a weak acid. To a greater or lesser extent, this behavior applies to other weak acids as well, such as HCN which dissociates into cyanide ions (\(CN^-\)) and hydronium ions.
Ionization Constant
The ionization constant (\(K_a\)) of a weak acid is a crucial parameter in determining the extent of dissociation of that acid in an aqueous solution. It is defined by the equilibrium expression: \[ K_a = \frac{[A^-][H_3O^+]}{[HA]} \] Where [A-] is the concentration of the conjugate base, [H3O+] is the concentration of hydronium ions and [HA] is the concentration of the undissociated acid.

The ionization constant is determined experimentally and allows for the prediction of the behavior of the acid in solution. A smaller Ka value indicates a weaker acid, meaning it donates protons less readily and dissociates less in solution. The precise determination of this constant is fundamental for the calculation of pH, as it is one of the key variables in the formula used to find the hydronium ion concentration for weak acids.
Hydronium Ion Concentration
The concentration of hydronium ions (\(H_3O^+\)) in an aqueous solution is directly related to its acidity. In the context of weak acid solutions, this concentration is not simply the original acid concentration due to the partial dissociation of the weak acid.

Using the ionization constant and the initial concentration of the acid, we can establish a quadratic equation where the unknown, typically represented as x, signifies the concentration of hydronium ions produced. This quantity, once calculated, is directly plugged into the pH formula. Therefore, understanding how to determine the hydronium ion concentration is key to pH calculation and to understanding the nature of acid-base chemistry. This fundamental concept also allows for further exploration into buffer solutions, acid-base titrations, and many other aspects of chemical equilibria.

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Most popular questions from this chapter

Trichloroacetic acid \(\left(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\) is a corrosive acid that is used to precipitate proteins. The \(\mathrm{pH}\) of a \(0.050-M\) solution of trichloroacetic acid is the same as the \(\mathrm{pH}\) of a \(0.040-\mathrm{M} \mathrm{HClO}_{4}\) solution. Calculate \(K_{\mathrm{a}}\) for trichloroacetic acid.

Making use of the assumptions we ordinarily make in calculating the \(\mathrm{pH}\) of an aqueous solution of a weak acid, calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-6}-M\) solution of hypobromous acid (HBrO, \(K_{\mathrm{a}}=2 \times 10^{-9}\) ). What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

Rank the following \(0.10 M\) solutions in order of increasing \(\mathrm{pH}\). a. HI, HF, NaF, NaI b. \(\mathrm{NH}_{4} \mathrm{Br}, \mathrm{HBr}, \mathrm{KBr}, \mathrm{NH}_{3}\) c. \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{NO}_{3}, \mathrm{NaNO}_{3}, \mathrm{NaOH}, \mathrm{HOC}_{6} \mathrm{H}_{5}, \mathrm{KOC}_{6} \mathrm{H}_{5}\), \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}, \mathrm{HNO}_{3}\)

Arsenic acid \(\left(\mathrm{H}_{3} \mathrm{AsO}_{4}\right)\) is a triprotic acid with \(K_{\mathrm{a}_{1}}=5.5 \times\) \(10^{-3}, K_{\mathrm{a}_{2}}=1.7 \times 10^{-7}\), and \(K_{\mathrm{a}_{3}}=5.1 \times 10^{-12}\). Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right],\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right],\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right],\left[\mathrm{HAsO}_{4}^{2-}\right]\), and \(\left[\mathrm{AsO}_{4}{ }^{3-}\right]\) in a \(0.20-M\) arsenic acid solution.

Calculate the \(\mathrm{pH}\) of a \(2.0-M \mathrm{H}_{2} \mathrm{SO}_{4}\) solution.

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