プリンキピア

第14章酸塩基平衡

Key Concepts and Summary重要な概念とまとめ

Concentrations of hydronium and hydroxide ions in aqueous media are often represented as logarithmic pH and pOH values, respectively. At 25 °C, the autoprotolysis equilibrium for water requires the sum of pH and pOH to equal 14 for any aqueous solution. The relative concentrations of hydronium and hydroxide ion in a solution define its status as acidic ([H3O+] > [OH−]), basic ([H3O+] < [OH−]), or neutral ([H3O+] = [OH−]). At 25 °C, a pH < 7 indicates an acidic solution, a pH > 7 a basic solution, and a pH = 7 a neutral solution.
水を溶媒とする系のオキソニウムイオン濃度と水酸化物イオン濃度は、それぞれ対数を用いたpHとpOHで表すことが多い。25 °Cでは、水の自己プロトン移動平衡により、どの水溶液でもpHとpOHの和は14となる。溶液中のオキソニウムイオンと水酸化物イオンの濃度の大小関係によって、酸性（[H3O+] > [OH−]）、塩基性（[H3O+] < [OH−]）、中性（[H3O+] = [OH−]）が決まる。25 °Cでは、pH < 7なら酸性溶液、pH > 7なら塩基性溶液、pH = 7なら中性溶液を示す。

Key Equations重要な式

pH=−log[H3O+]
pOH = −log[OH−]
[H3O+] = 10−pH
[OH−] = 10−pOH
pH + pOH = pKw = 14.00 at 25 °C
pH + pOH = pKw = 14.00（25 °C）

Chemistry End of Chapter Exercises章末問題

Exercises 15–25練習問題 15–25
Exercise 15問題 15
Explain why a sample of pure water at 40 °C is neutral even though [H3O+] = 1.7 × 10−7 M. Kw is 2.9 × 10−14 at 40 °C.
40 °Cの純水が、[H3O+] = 1.7 × 10−7 Mであるにもかかわらず中性である理由を説明しなさい。40 °CでのKwは2.9 × 10−14である。
Solution解き方
In a neutral solution [H3O+] = [OH−]. At 40 °C, [H3O+] = [OH−] = (2.910 × 10−14)1/2 = 1.7 × 10−7.
中性溶液では[H3O+] = [OH−]である。40 °Cでは、[H3O+] = [OH−] = (2.910 × 10−14)1/2 = 1.7 × 10−7。
Exercise 16問題 16
The ionization constant for water (Kw) is 2.9 × 10−14 at 40 °C. Calculate [H3O+], [OH−], pH, and pOH for pure water at 40 °C.
水の解離定数（Kw）は40 °Cで2.9 × 10−14である。40 °Cの純水の[H3O+]、[OH−]、pH、pOHを計算しなさい。
Exercise 17問題 17
The ionization constant for water (Kw) is 9.311 × 10−14 at 60 °C. Calculate [H3O+], [OH−], pH, and pOH for pure water at 60 °C.
水の解離定数（Kw）は60 °Cで9.311 × 10−14である。60 °Cの純水の[H3O+]、[OH−]、pH、pOHを計算しなさい。
Solution解き方
x = 3.051 × 10−7 M = [H3O+] = [OH−]; pH = −log3.051 × 10−7 = −(−6.5156) = 6.5156; pOH = pH = 6.5156
x = 3.051 × 10−7 M = [H3O+] = [OH−]；pH = −log3.051 × 10−7 = −(−6.5156) = 6.5156；pOH = pH = 6.5156
Exercise 18問題 18
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
溶質が完全に電離する、次の各溶液の25 °CでのpHとpOHを計算しなさい。
(a) 0.200 M HCl
(a) 0.200 M HCl
(b) 0.0143 M NaOH
(b) 0.0143 M NaOH
(c) 3.0 M HNO3
(c) 3.0 M HNO3
(d) 0.0031 M Ca(OH)2
(d) 0.0031 M Ca(OH)2
Exercise 19問題 19
Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
溶質が完全に電離する、次の各溶液の25 °CでのpHとpOHを計算しなさい。
(a) 0.000259 M HClO4
(a) 0.000259 M HClO4
(b) 0.21 M NaOH
(b) 0.21 M NaOH
(c) 0.000071 M Ba(OH)2
(c) 0.000071 M Ba(OH)2
(d) 2.5 M KOH
(d) 2.5 M KOH
Solution解き方
(a) pH = 3.587; pOH = 10.413; (b) pOH = 0.68; pH = 13.32; (c) pOH = 3.85; pH = 10.15; (d) pOH = −0.40; pH = 14.4
(a) pH = 3.587; pOH = 10.413; (b) pOH = 0.68; pH = 13.32; (c) pOH = 3.85; pH = 10.15; (d) pOH = −0.40; pH = 14.4
Exercise 20問題 20
What are the pH and pOH of a solution of 2.0 M HCl, which ionizes completely?
完全に電離する2.0 M HCl水溶液のpHとpOHはいくらか。
Exercise 21問題 21
What are the hydronium and hydroxide ion concentrations in a solution whose pH is 6.52?
pHが6.52の溶液のオキソニウムイオン濃度と水酸化物イオン濃度はいくらか。
Solution解き方
[H3O+] = 3.0 × 10−7 M; [OH−] = 3.3 × 10−8 M
[H3O+] = 3.0 × 10−7 M; [OH−] = 3.3 × 10−8 M
Exercise 22問題 22
Calculate the hydrogen ion concentration and the hydroxide ion concentration in wine from its pH. See Figure 14.2 for useful information.
ワインのpHから、水素イオン濃度と水酸化物イオン濃度を計算しなさい。必要な情報は図14.2を参照。
Exercise 23問題 23
Calculate the hydronium ion concentration and the hydroxide ion concentration in lime juice from its pH. See Figure 14.2 for useful information.
ライム果汁のpHから、オキソニウムイオン濃度と水酸化物イオン濃度を計算しなさい。必要な情報は図14.2を参照。
Solution解き方
[H3O+] = 1 × 10−2 M; [OH−] = 1 × 10−12 M
[H3O+] = 1 × 10−2 M; [OH−] = 1 × 10−12 M
Exercise 24問題 24
The hydronium ion concentration in a sample of rainwater is found to be 1.7 × 10−6 M at 25 °C. What is the concentration of hydroxide ions in the rainwater?
ある雨水のオキソニウムイオン濃度は、25 °Cで1.7 × 10−6 Mであるとわかった。この雨水の水酸化物イオン濃度はいくらか。
Exercise 25問題 25
The hydroxide ion concentration in household ammonia is 3.2 × 10−3 M at 25 °C. What is the concentration of hydronium ions in the solution?
家庭用アンモニア水の水酸化物イオン濃度は、25 °Cで3.2 × 10−3 Mである。この溶液のオキソニウムイオン濃度はいくらか。
Solution解き方
[OH−] = 3.1 × 10−12 M
[OH−] = 3.1 × 10−12 M
acidic
酸性
a solution in which [H3O+] > [OH−]
[H3O+] > [OH−]である溶液
basic
塩基性
a solution in which [H3O+] < [OH−]
[H3O+] < [OH−]である溶液
neutral
中性
describes a solution in which [H3O+] = [OH−]
[H3O+] = [OH−]である溶液を表す
pH
pH
logarithmic measure of the concentration of hydronium ions in a solution
溶液中のオキソニウムイオン濃度を対数で表した尺度
pOH
pOH
logarithmic measure of the concentration of hydroxide ions in a solution
溶液中の水酸化物イオン濃度を対数で表した尺度