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A student mistakenly calculates the \(\mathrm{pH}\) of a \(1.0 \times 10^{-7} \mathrm{M}\) HI solution to be \(7.0 .\) Explain why the student is incorrect and calculate the correct pH.

Short Answer

Expert verified
The student is incorrect because they neglected the contribution of H+ ions from water autoionization. The correct pH calculation includes this contribution, giving a pH of 6.70.

Step by step solution

01

Understand the Concept of pH and the Ionization of Strong Acids

The pH of a solution is the negative logarithm of the hydrogen ion concentration. Since HI is a strong acid, it completely dissociates in water. Therefore, a 1.0 x 10^-7 M solution of HI should yield 1.0 x 10^-7 M of H+ ions. However, we must consider the contribution of H+ ions from water autoionization as well.
02

Consider Autoionization of Water

Water itself undergoes autoionization, producing 1.0 x 10^-7 M of H+ ions at 25 degrees Celsius. Because additional H+ ions come from water, the total H+ concentration will be greater than 1.0 x 10^-7 M, resulting in a pH less than 7.0.
03

Calculate the Total [H+] Concentration

The total concentration of H+ ions is the sum from the acid and water: total [H+] = [H+ from HI] + [H+ from water] = 1.0 x 10^-7 M + 1.0 x 10^-7 M = 2.0 x 10^-7 M.
04

Calculate the Correct pH

Use the pH formula: pH = -log[H+] = -log(2.0 x 10^-7). Calculate the pH using a calculator.
05

Finalize the Answer

After calculation, if you obtain pH = 6.70, this is the correct pH of the 1.0 x 10^-7 M HI solution.

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Key Concepts

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

Autoionization of Water
The autoionization of water is a fundamental concept in acid-base chemistry. It refers to the process where water molecules, H2O, naturally split into hydrogen ions (H+) and hydroxide ions (OH-). At 25 degrees Celsius, this phenomenon is in a delicate balance, resulting in a constant product, known as the water ionization constant (Kw), which has a value of approximately 1.0 x 10-14 at standard conditions.

When water autoionizes, equal concentrations of H+ and OH- ions are produced. Hence, pure water has a concentration of about 1.0 x 10-7M for both ions. This equilibrium must be considered when calculating the pH of solutions, since any additional source of H+ ions, like a strong acid, will shift the balance. Incorporating the effects of autoionization typically means that even the pH of very dilute strong acid solutions will be less than 7, indicating that the solution is still acidic due to the presence of extra H+ ions.
Ionization of Strong Acids
Strong acids fully dissociate in water, releasing a corresponding amount of hydrogen ions (H+). This process is termed ionization. Acids like hydroiodic acid (HI), hydrochloric acid (HCl), and sulfuric acid (H2SO4) are considered strong acids because of their complete ionization in aqueous solutions.

The total [H+] concentration of an aqueous solution, with a strong acid like HI, will be the sum of the [H+] from the dissolved acid and from the water's autoionization. Since these acids ionize completely, the concentration of H+ ions released by them will be equal to the initial concentration of the acid before it is dissolved in water. Therefore, it is essential in pH calculations to remember that the complete ionization of strong acids contributes significantly to the overall hydrogen ion concentration.
Negative Logarithm of Hydrogen Ion Concentration
pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm (base 10) of the hydrogen ion concentration, symbolized as [H+]. The pH scale typically ranges from 0 to 14, with 7 being neutral; pH values less than 7 indicate acidity, and values greater than 7 indicate alkalinity.

Mathematically, the pH is given by the formula: pH = -log[H+]. This logarithmic relationship means that as the hydrogen ion concentration increases by a factor of 10, the pH decreases by 1 unit. With logarithms being less intuitive to grasp, students should practice calculating pH using different hydrogen ion concentrations to understand its scale. Also, it's noteworthy that due to the nature of the log function, small changes in [H+] can result in significant changes in pH, reflecting a wide range of acidity or basicity within solutions.

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Most popular questions from this chapter

Acid Strength and \(K_{\mathrm{a}}\) Classify each acid as strong or weak. If the acid is weak, write an expression for the acid ionization constant \(\left(K_{a}\right) .\) $$\begin{array}{l}{\text { a. HF }} \\ {\text { b. HCHO }} \\ {\text { c. } \mathrm{H}_{2} \mathrm{SO}_{4}} \\ {\text { d. } \mathrm{H}_{2} \mathrm{CO}_{3}}\end{array}$$

Acid Strength and \(K_{\mathrm{a}}\) Rank the solutions in order of decreasing \(\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] : 0.10 \mathrm{M} \mathrm{HCl} ; 0.10\) \(\mathrm{M} \mathrm{HF} ; 0.10 \mathrm{MClO} ; 0.10 \mathrm{MHC}_{6} \mathrm{H}_{5} \mathrm{O} .\)

What is the Lewis definition of an acid? Of a base?

In each reaction, identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid, and the conjugate base. $$\begin{array}{l}{\text { a. } \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q)} \\ {\text { b. } \mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{NH}_{4}^{+}(a q)+\mathrm{OH}^{-}(a q)} \\ {\text { c. } \mathrm{HNO}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q)} \\ {\text { d. } \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{OH}^{-}(a q)}\end{array}$$

Determine the \(\mathrm{pH}\) of each two-component solution. $$\begin{array}{l}{\text { a. } 0.050 \mathrm{M} \mathrm{KOH} \text { and } 0.015 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}} \\ {\text { b. } 0.265 \mathrm{M} \mathrm{NH}_{4} \mathrm{NO}_{3} \text { and } 0.102 \mathrm{M} \mathrm{HCN}} \\ {\text { c. } 0.075 \mathrm{M} \mathrm{RbOH} \text { and } 0.100 \mathrm{M} \mathrm{NaHCO}_{3}} \\ {\text { d. } 0.088 \mathrm{M} \mathrm{HClO}_{4} \text { and } 0.022 \mathrm{M} \mathrm{KOH}} \\ {\text { e. } 0.115 \mathrm{M} \mathrm{NaClO} \text { and } 0.0500 \mathrm{M} \mathrm{KI}}\end{array}$$

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