Calculate the pH of each of the following solutions. a. 0.10 M C H3 N H3 C I b. 0.050 M NaCN

Name: Problem 119, Chapter 13: Chemistry An Atoms First Approach
Uploaded: 2020-08-31T19:21:20-08:00
Duration: 5 min 23 s
Channel: Ronald Prasad
Description: Calculate the pH of each of the following solutions. a. 0.10 M C H3 N H3 C I b. 0.050 M NaCN

step 1 Let's calculate the pH of each of the following solutions. For A, we have 0 .10 molar, CH3, NH3

Calculate the pH of each of the following solutions. a. 0.10...

Key Concepts

Conjugate Acid?Base Pairs

This concept deals with the relationship between a weak acid and its conjugate base (or a weak base and its conjugate acid). In aqueous solutions, salts formed from weak acids or bases dissociate completely and produce conjugate species that can react with water. In the case of CH3NH3Cl, the cation (CH3NH3+) is the conjugate acid of a weak base, while for NaCN the anion (CN–) is the conjugate base of a weak acid. The behavior of these conjugate species is central to understanding how they affect the pH of the solution.

Salt Hydrolysis

Salt hydrolysis refers to the reaction of the ions produced from a dissolved salt with water, which can lead to the production of either hydronium ions (H3O+) or hydroxide ions (OH–). The resulting equilibrium affects the pH of the solution. For example, in an aqueous solution of a salt derived from a weak base (such as CH3NH3Cl), the cation hydrolyzes to form H3O+, making the solution acidic. Conversely, a salt derived from a weak acid (like NaCN) will have its anion react with water to generate OH–, resulting in a basic solution.

Equilibrium Constant Relationships (Ka, Kb, Kw)

This principle connects the acid dissociation constant (Ka) and the base dissociation constant (Kb) of conjugate pairs through the ion product of water (Kw). When one constant is known (for either the weak acid or weak base), the other can be calculated using the relation Ka × Kb = Kw. This relationship is pivotal in determining the extent of hydrolysis of the conjugate species and, hence, the pH of the solution.

pH Calculation in Weak Acid/Base Systems

Calculating the pH of solutions involving weak acids, weak bases, or their salts typically requires setting up equilibrium expressions that incorporate the relevant dissociation or hydrolysis constants. Approximate methods, such as using the formula pH = ½(pKa - log C) for weak acids (or its counterpart for weak bases), can simplify these calculations. Understanding these approximations and how they arise from equilibrium considerations is essential for estimating the pH of solutions produced by salts of weak acids or bases.

Transcript

00:01 Let's calculate the ph of each of the following solutions.

00:03 For a, we have 0 .10 molar, ch3, nh3 cl.

00:12 Let's first break apart the compound here to ch3nh3 plus and cl minus.

00:25 This is a weak acid, and we can write our equilibrium, c .h3nh3 plus and h2o in equilibrium with c .h3nh2 and h3o plus.

00:46 This is 0 .10 molar.

00:49 This is 0 .10 molar, which makes the initial molarity here.

00:54 0 .10 molar.

00:56 Let's complete our ice table.

01:04 Ka is equal to h3o.

01:08 Plus ch3 nh2 over c .h3 nh3 plus.

01:18 And looking up the ka here, we need to calculate the ka.

01:29 The kb for ch3 methylamine nh2 2 is equal to 4 .38 times 10 to the negative 4.

01:54 Therefore, the k -a for the conjugate, ch3nh3 plus, would we call the 1 .0 times 10 to the negative 1 .00 times 10 to negative 14, over 4 .38 times 10 to the negative 4.

02:08 And working out the ka for the conjugate here, yields 2 .28 times 10 in the negative 11.

02:15 Let's bring that over here.

02:16 2 .28 times 10 to the negative 11.

02:19 It's equal to x, x and 0 .10 minus x.

02:26 We are going to solve this equation here, and solving for x will yield 1 .50 times 10 to the negative 6...

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