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16.7:

滴定計算:弱酸-弱塩基

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JoVE Core Chemistry
Titration Calculations: Weak Acid – Strong Base

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The pH at different stages of a weak acid or base titration is calculated using different methods during various points of the titration. If a weak acid or base is the main determinant of the pH, the Ka or Kb and an ICE table, or the Henderson-Hasselbalch equation are used. If strong acid or base is present after the neutralization reaction, the concentration of the remaining hydronium or hydroxide ions is used to calculate the pH. The initial pH of 50 mL of a 0.10 M acetic acid solution is 2.87 and is calculated using the Ka and an ICE table, as acetic acid is the main contributor. If this solution containing 0.0050 moles of acetate is titrated with 0.10 M sodium hydroxide, the hydroxide ions react with acetic acid to produce acetate, resulting in a buffer. Therefore, when 10 mL of the sodium hydroxide containing 0.0010 moles of hydroxide ions is added, 0.0010 moles of acetate are formed, and 0.0040 moles of acetic acid remain. The pH of the buffer can be calculated by substituting these values into the Henderson-Hasselbalch equation and equals 4.14. When 25 mL of the sodium hydroxide is added, half of the initial moles of acetic acid are converted into acetate. At this point, the pH equals the pKa, as the amount of acetic acid and acetate ion are equal. The further addition of sodium hydroxide up to 50 mL converts all of the acetic acid molecules into acetate, and the equivalence point is reached. As acetate ions are basic, the equivalence point lies in the basic region. The concentration of the acetate ion is calculated by dividing the number of moles by the total volume of the solution.  The pH is determined using the Kb for acetate ions and an ICE table, as the acetate ion is the main contributor to the pH at the equivalence point. The Kb for acetate is calculated using the formula, Kw = Ka × Kb, and equals 5.6 × 10−10. Substituting equilibrium concentrations in the expression for Kb gives the hydroxide concentration, 5.3 × 10−6 M. The pOH and pH of the solution are 5.28 and 8.72, respectively.  Further addition of sodium hydroxide in the solution results in a mixture of acetate ions and sodium hydroxide. However, the final concentration of sodium hydroxide determines the pH of the solution, as it is a stronger base than acetate. Therefore, if 70 mL of sodium hydroxide is added into the solution, the final concentration of the hydroxide ions can be calculated by subtracting the total moles of acetic acid, 0.0050 moles, from the total moles of hydroxide ions added, 0.0070 moles, and dividing it by the total volume of the solution, 120 mL or 0.12 L.   As the hydroxide ion concentration is 0.017 M, the pOH and pH of the solution are calculated to be 1.78 and 12.22, respectively.

16.7:

滴定計算:弱酸-弱塩基

滴定溶液のpH計算: 弱酸および強塩基

25.00 mLの0.100 M CH3CO2Hを0.100 M NaOHで滴定する場合、反応は次のように表すことができます。

Eq1

異なる容量のNaOH滴定液を添加した後の滴定液のpHは、以下のように計算できます。

(a) 酢酸溶液の初期pHは、通常のICE法で計算されます。

Eq2

(b) 酸と滴定試薬はともに単分子から構成され、試料と滴定試薬は等濃度であるため、この滴定試薬の量が当量点を表します。強酸性の例とは異なり、この場合の反応混合物には弱い共役塩基(酢酸イオン)が含まれています。以下の濃度で存在する酢酸の塩基イオン化を考慮して、溶液のpHを計算します。

Eq3

酢酸のイオン化は、次の式で表されます。

Eq4

x << 0.0500とすると、pHは通常のICE法で計算できます。

Eq5

弱酸と強塩基を滴定したときに予想されるように、この滴定の当量点でのpHは7よりもかなり大きいことに注意する必要があります。

(c) 滴定溶液 = 12.50 mLの場合。この量は、化学量論的な滴定液量の2分の1に相当し、2分の1の酢酸が中和されて、同量の酢酸イオンが生成します。すなわち、酸と共役塩基の濃度は等しいです。この場合、ヘンダーソン・ハッセルバルヒ式を用いてpHを計算するのが便利です。

Eq6

(pH = 弱酸性の滴定における半当量点でのpKa)

(d) 滴定液量 = 37.50 mLの場合。この量は化学量論的に過剰であり、生成物である酢酸イオンと過剰量の滴定試薬の両方を含む反応溶液となります。このような溶液では、溶液のpHは主に過剰量の強塩基により決まります。

Eq7

上記の文章は以下から引用しました。 Openstax, Chemistry 2e, Section 14.7: Acid-base Titrations.