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Chapter 13: Problem 57

Calculate the \(\mathrm{pH}\) of each of the following solutions of a strong acid in water. a. \(0.10 \mathrm{M} \mathrm{HCl}\) b. 5.0 M HCl c. \(1.0 \times 10^{-11} \mathrm{M} \mathrm{HCl}\)

Short Answer

Expert verified
The pH of each solution is as follows: a. pH of 0.10 M HCl is 1, b. pH of 5.0 M HCl is approximately -0.70, and c. pH of \(1.0 \times 10^{-11}\) M HCl is 11.

Step by step solution

01

(Part a: Calculate [H₃O+] for 0.10 M HCl solution)

For a strong acid like HCl, the concentration of H₃O+ ions is equal to the concentration of HCl as it dissociates completely. So, in this case, [H₃O+] = 0.10 M.
02

(Part a: Calculate pH of 0.10 M HCl solution)

Now we will use the pH formula: pH = -log[H₃O+] Plugging in the value of [H₃O+] = 0.10 M, we get: pH = -log(0.10) = 1 The pH of the 0.10 M HCl solution is 1.
03

(Part b: Calculate [H₃O+] for 5.0 M HCl solution)

In this case as well, the concentration of H₃O+ ions is equal to the concentration of HCl, which is 5.0 M.
04

(Part b: Calculate pH of 5.0 M HCl solution)

Using the pH formula, we have: pH = -log[H₃O+] pH = -log(5.0) ≈ -0.70 The pH of the 5.0 M HCl solution is approximately -0.70.
05

(Part c: Calculate [H₃O+] for \(1.0 \times 10^{-11}\) M HCl solution)

Again, the concentration of H₃O+ ions is equal to the concentration of HCl, which is \(1.0 \times 10^{-11}\) M.
06

(Part c: Calculate pH of \(1.0 \times 10^{-11}\) M HCl solution)

Using the pH formula, we get: pH = -log[H₃O+] pH = -log(\(1.0 \times 10^{-11}\)) = 11 The pH of the \(1.0 \times 10^{-11}\) M HCl solution is 11.

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Most popular questions from this chapter

A \(0.20-M\) sodium chlorobenzoate \(\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution has a pH of \(8.65 .\) Calculate the pH of a 0.20- \(M\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.

Consider 0.10 \(M\) solutions of the following compounds: \(\mathrm{AlCl}_{3}, \mathrm{NaCN}, \mathrm{KOH}, \mathrm{CsClO}_{4},\) and NaF. Place these solutions in order of increasing \(\mathrm{pH}\).

Calculate the \(\mathrm{pH}\) of a \(0.200-M\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The principal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is $$\begin{aligned} \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \\ \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q) &+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\end{aligned}$$

Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4},\) is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise 146.) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: \(\mathrm{CO}_{2}\) blood levels increase during cardiac arrest.

Place the species in each of the following groups in order of increasing acid strength. Explain the order you chose for each group. a. \(\mathrm{HIO}_{3}, \mathrm{HBrO}_{3}\) b. \(\mathrm{HNO}_{2}, \mathrm{HNO}_{3}\) c. HOCI, HOI d. \(\mathrm{H}_{3} \mathrm{PO}_{4}, \mathrm{H}_{3} \mathrm{PO}_{3}\)
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