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Chapter 14: Problem 57

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

Short Answer

Expert verified
The pH of the given solutions are: a) For the \(0.10 M\) HCl solution, the pH is 1. c) For the \(1.0 \times 10^{-11} M\) HCl solution, the pH is 11. b) For the \(5.0 M\) HCl solution, the pH is approximately -0.3.

Step by step solution

01

Determine the concentration of H+ ions

Since HCl is a strong acid, it will dissociate completely in water. So, the concentration of H+ ions will be equal to the concentration of HCl in the solution. For this solution, [H+] = \([0.10 M]\).
02

Calculate the pH of the solution

To calculate the pH, use the formula: pH = -log[H+]. So, for this solution, pH = - log\[0.10\] = 1. For solution c:
03

Determine the concentration of H+ ions

Since HCl is a strong acid, it will dissociate completely in water. So, the concentration of H+ ions will be equal to the concentration of HCl in the solution. For this solution, [H+] = \(1.0 \times 10^{-11} M\).
04

Calculate the pH of the solution

To calculate the pH, use the formula: pH = -log[H+]. So, for this solution, pH = - log\[\(1.0 \times 10^{-11}\)\] = 11. For solution b:
05

Determine the concentration of H+ ions

Since HCl is a strong acid, it will dissociate completely in water. So, the concentration of H+ ions will be equal to the concentration of HCl in the solution. For this solution, [H+] = \([5.0 M]\).
06

Calculate the pH of the solution

To calculate the pH, use the formula: pH = -log[H+]. So, for this solution, pH = - log\[5.0\]. Since 5.0 is between 1 and 10, the pH will be between 0 and 1. We can further estimate it as pH ≈ -0.3.

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Most popular questions from this chapter

The \(\mathrm{pH}\) of \(1.0 \times 10^{-8} \mathrm{M}\) hydrochloric acid is not \(8.00\). The correct \(\mathrm{pH}\) can be calculated by considering the relationship between the molarities of the three principal ions in the solution \(\left(\mathrm{H}^{+}, \mathrm{Cl}^{-}\right.\), and \(\left.\mathrm{OH}^{-}\right) .\) These molarities can be calculated from algebraic equations that can be derived from the considerations given below. a. The solution is electrically neutral. b. The hydrochloric acid can be assumed to be \(100 \%\) ionized. c. The product of the molarities of the hydronium ions and the hydroxide ions must equal \(K_{\mathrm{w}}\). Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-8}-\mathrm{M} \mathrm{HCl}\) solution.

Sodium azide \(\left(\mathrm{NaN}_{3}\right)\) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a \(0.010-M\) solution of \(\mathrm{NaN}_{3}\). The \(K_{\mathrm{a}}\) value for hydrazoic acid \(\left(\mathrm{HN}_{3}\right)\) is \(1.9 \times 10^{-5}\)

A codeine-containing cough syrup lists codeine sulfate as a major ingredient instead of codeine. The Merck Index gives \(\mathrm{C}_{36} \mathrm{H}_{44} \mathrm{~N}_{2} \mathrm{O}_{10} \mathrm{~S}\) as the formula for codeine sulfate. Describe the composition of codeine sulfate. (See Exercise 155.) Why is codeine sulfate used instead of codeine?

Calculate the \(\mathrm{pH}\) of a \(1.0 \times 10^{-7}-M\) solution of \(\mathrm{NaOH}\) in water.

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