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Chapter 13: Problem 100

Calculate the \(\mathrm{pH}\) of a \(0.050-M\left(\mathrm{C}_{2} \mathrm{H}_{5}\right)_{2} \mathrm{NH}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(\left.1.3 \times 10^{-3}\right)\)

Short Answer

Expert verified
The pH of the 0.050 M diethylamine \((C_2H_5)_2NH\) solution is 10.51.

Step by step solution

01

Write the acid-base reaction and the Kb expression

The reaction for the dissociation of diethylamine (C2H5)2NH is given by: \((C_2H_5)_2NH+H_2O \Leftrightarrow (C_2H_5)_2NH_2^+ + OH^-\) Now, we can write the Kb expression based on the given value: \(K_b = \frac{[(C_2H_5)_2NH_2^+][OH^-]}{[(C_2H_5)_2NH]}\) We know that the initial concentration of diethylamine is 0.050 M, and initially, the concentrations of (C2H5)2NH2+ and OH- are 0.
02

Create an ICE table

To determine the concentrations of the species at equilibrium, set up an ICE table: \[\begin{array}{cccc} & -x & -x & +x \\ (C_2H_5)_2NH & 0.050 & 0 & 0 \\ H_2O & - & - & - \\ (C_2H_5)_2NH_2^+ & 0 & +x & \\ OH^- & 0 & +x & \\ \end{array}\]
03

Substitute the equilibrium concentrations into the Kb expression

Now, we have the following algebraic equation for x: \(K_b = \frac{x^2}{0.050 - x}\)
04

Solve for x

Since Kb is small, we can approximate x as being much smaller than 0.050, so: \(K_b = \frac{x^2}{0.050}\) Solving for x, we get: \(x = [\ce{OH^{-}}] = \sqrt{0.050 * K_b} = \sqrt{0.050 * 1.3 * 10^{-3}} = 3.22 * 10^{-4} M\)
05

Calculate pOH

Now that we have found the concentration of OH-, we can calculate the pOH using the formula: \(pOH = -\log_{10} [\ce{OH^{-}}]\) \(pOH = -\log_{10} (3.22 * 10^{-4}) = 3.49\)
06

Calculate pH

Finally, we can find the pH using the relationship between pH, pOH, and the ion product of water (Kw): \(pH + pOH = 14\) Thus, the pH is: \(pH = 14 - pOH = 14 - 3.49 = 10.51\) So, the pH of the given diethylamine solution is 10.51.

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Most popular questions from this chapter

Classify each of the following as a strong acid, weak acid, strong base, or weak base in aqueous solution. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{HNO}_{3}\) c. \(\mathrm{CH}_{3} \mathrm{NH}_{2}\) d. \(\mathrm{NaOH}\) e. \(\mathrm{NH}_{3}\) f. HF g. h. \(\mathrm{Ca}(\mathrm{OH})_{2}\) i. \(\mathrm{H}_{2} \mathrm{SO}_{4}\)

Arrange the following 0.10 \(M\) solutions in order of most acidic to most basic. \(\begin{array}{llll} & \text { KOH, } & \text { KNO }_{3}, & \text { KCN, } & \text { NH_Cl, } & \text { HCl }\end{array}\)

Calculate the \(\mathrm{pH}\) of each of the following solutions. a. \(0.12 M \mathrm{KNO}_{2}\) b. \(0.45 M\) NaOCl c. \(0.40 M \mathrm{NH}_{4} \mathrm{ClO}_{4}\)

A solution of formic acid (HCOOH, \(K_{\mathrm{a}}=1.8 \times 10^{-4}\) ) has a pH of \(2.70 .\) Calculate the initial concentration of formic acid in this solution.

Write out the stepwise \(K_{\mathrm{a}}\) reactions for citric acid \(\left(\mathrm{H}_{3} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{O}_{7}\right)\) a triprotic acid.
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