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Chapter 14: Problem 16

Why is the pH of water at \(25^{\circ} \mathrm{C}\) equal to 7.00\(?\)

Short Answer

Expert verified
At 25掳C, the autoionization of water results in equal concentrations of hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)), with the ion product constant \(K_w = 1.0 \times 10^{-14} M^2\). The concentration of hydrogen ions in pure water at this temperature is \(1.0 \times 10^{-7}\) M. Since pH is defined as the negative logarithm (base 10) of the hydrogen ion concentration, the pH of water at 25掳C is: \[pH = -\log_{10} (1.0 \times 10^{-7}) = 7.00\]

Step by step solution

01

Understanding pH

pH is a measure of how acidic or basic a solution is. It is defined as the negative logarithm (base 10) of the hydrogen ion concentration (\([H^+]\)) in moles per liter (M). In mathematical terms, pH is given as: \[pH = -\log_{10} [H^+]\]
02

Autoionization of water

At 25掳C, water undergoes autoionization, a process in which water molecules split into hydrogen ions (\(H^+\)) and hydroxide ions (\(OH^-\)). This can be represented by the following equilibrium equation: \[H_2O \rightleftharpoons H^+ + OH^-\] Under these conditions, the concentration of hydrogen ions and hydroxide ions in pure water is equal, and their product (\([H^+][OH^-]\)) equals the ion product constant for water (\(K_w\)). At 25掳C, \(K_w\) is equal to \(1.0 \times 10^{-14} M^2\).
03

Calculating the concentration of hydrogen ions

Since the concentrations of hydrogen and hydroxide ions are equal in pure water at 25掳C, we can write the ion product constant equation as: \[[H^+][OH^-] = K_w\] Substitute the value of \(K_w\) and solve for the concentration of hydrogen ions (\([H^+]\)): \[[H^+][H^+] = 1.0 \times 10^{-14}\] \[[H^+]^2 = 1.0 \times 10^{-14}\] \[[H^+] = \sqrt{1.0 \times 10^{-14}}\] \[[H^+] = 1.0 \times 10^{-7}\]
04

Calculating the pH of water

We now know that the hydrogen ion concentration in water at 25掳C is \(1.0 \times 10^{-7}\) M. To find the pH value, we use the pH formula: \[pH = -\log_{10} [H^+]\] Substitute the value of \([H^+]\) and solve for pH: \[pH = -\log_{10} (1.0 \times 10^{-7})\] \[pH = 7.00\] So, at 25掳C, the pH of pure water is equal to 7.00.

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Key Concepts

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

Autoionization of Water
Understanding the concept of autoionization is essential to grasp why water, even though it appears neutral, can act as both an acid and a base. Autoionization is a self-ionization process where water molecules dissociate into hydrogen ions \( H^+ \) and hydroxide ions \( OH^- \). This can be represented by the equilibrium equation:\[ H_2O ightleftharpoons H^+ + OH^- \]This intriguing behavior happens because water is a polar molecule, meaning that its slightly positive hydrogen atoms and slightly negative oxygen atoms can interact with each other and split apart.
  • This self-ionization occurs naturally in all water samples.
  • Even in pure water, this reaction constantly reaches its equilibrium.
Typically, only a tiny fraction of water molecules undergo autoionization, but it's enough to produce measurable amounts of hydrogen and hydroxide ions.
Hydrogen Ion Concentration
Hydrogen ion concentration is crucial in determining a solution's pH. In a neutral solution at 25掳C, like pure water, the concentration of hydrogen ions \([H^+]\) is exactly equal to the concentration of hydroxide ions \([OH^-]\). This balance is due to the autoionization of water. When pure water is at 25掳C, it has about \(1.0 \times 10^{-7} \) moles per liter (M) of hydrogen ions.
This balance is what gives pure water its characteristic pH of 7.00.
  • The lower the \([H^+]\), the more basic the solution.
  • The higher the \([H^+]\), the more acidic the solution becomes.
It鈥檚 interesting to consider that even in a seemingly neutral substance like water, there's a constant and significant presence of ions affecting its chemical properties.
Ion Product Constant (Kw)
The ion product constant, known as \( K_w \), is a fundamental value in understanding aqueous solutions' chemistry. It refers to the product of the concentrations of hydrogen ions and hydroxide ions in water, expressed as:\[ K_w = [H^+][OH^-] \]At 25掳C, \( K_w \) for water is \(1.0 \times 10^{-14} M^2\). This constant is a benchmark for determining how acidic or basic a solution is.
Since \( K_w \) is established by the natural autoionization at this temperature, it reflects:
  • The intrinsic properties of water under standard conditions.
  • Shows that small ionic concentrations can significantly influence pH.
Understanding \( K_w \) helps illustrate how water can play both roles of acids and bases in a vast range of chemical reactions. With this knowledge, students can appreciate how minute changes in ion concentrations can lead to significant changes in pH.

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Most popular questions from this chapter

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

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C} ) .\) Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14} .\)

Calculate the \(\mathrm{pH}\) of a \(0.20-M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) solution \(\left(K_{\mathrm{b}}=\right.\) \(5.6 \times 10^{-4} )\)

Isocyanic acid \((\mathrm{HNCO})\) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation $$ 2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) $$ Upon isolating pure HNCO \((l),\) an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a 100 -mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and \(\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},\) assuming all of the HNCO produced is dissolved in solution? \(\left(K_{\mathrm{a}} \text { of HNCO }\right.\) \(=1.2 \times 10^{-4} . )\)

A \(1.0 \times 10^{-2}-M\) solution of cyanic acid (HOCN) is 17\(\%\) dissociated. Calculate \(K_{\mathrm{a}}\) for cyanic acid.
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