U8.7 · pH and pKaPractice · pH 与 pKa
Lessons covered:U8.7 pH and pKaU8.9 Henderson-Hasselbalch Equation
- 1Multiple choice
HNO₂ has pKa = 3.34. In a solution at pH 5.00, which form of nitrous acid predominates?
ExplanationWhen pH > pKa the base form predominates. [NO₂⁻]/[HNO₂] = 105.00 − 3.34 = 101.66 ≈ 46. The ratio depends only on the pH and pKa, not on the total concentration.
Review: U8.7 pH and pKa
- 2Numeric
HCN has pKa = 9.31. In a solution at pH 9.00, what percent of the cyanide species (HCN + CN⁻) is HCN?
%Explanation[CN⁻]/[HCN] = 109.00 − 9.31 = 10−0.31 = 0.49. The fraction that is HCN = 1 ÷ (1 + 0.49) = 0.67, about 67%. The pH is a little below the pKa, so the acid form is slightly ahead.
Review: U8.7 pH and pKa
- 3Multiple choice
Bromothymol blue is an indicator whose acid form HIn is yellow and base form In⁻ is blue, with a color change range of about pH 6.0–7.6. What color is it in a solution at pH 9.0?
ExplanationAn indicator is a weak acid, HIn ⇌ H⁺ + In⁻. At a pH about 2 units above its pKa (about 7), [In⁻]/[HIn] ≈ 100, so it is almost all blue In⁻. Beyond the range the color simply stops changing and shows that end’s color.
Review: U8.7 pH and pKa
- 4Multiple choice
A weak acid solution has a pH 2.0 units above the acid’s pKa. What is [A⁻] ÷ [HA]?
Explanationlog([A⁻]/[HA]) = pH − pKa = 2.0, so the ratio = 10² = 100. Each unit of pH above the pKa multiplies the base-to-acid ratio by 10; two units below the pKa the ratio is 0.01.
Review: U8.7 pH and pKa
- 5Multiple choice
Carbonic acid is diprotic: pKa₁ = 6.37 (H₂CO₃ ⇌ H⁺ + HCO₃⁻) and pKa₂ = 10.33 (HCO₃⁻ ⇌ H⁺ + CO₃²⁻). In a solution at pH 8.50, which carbon species predominates?
ExplanationpH 8.50 is above pKa₁ (6.37), so most H₂CO₃ has become HCO₃⁻, and below pKa₂ (10.33), so little HCO₃⁻ has gone on to CO₃²⁻. HCO₃⁻ predominates. Between the two pKa values, the middle form wins.
Review: U8.7 pH and pKa
- 6Multiple choice
The graph shows the fractions of acetic acid present as and as against pH. At pH 6.0, about what percent of the acetic acid is present as ?
ExplanationThe curves cross at pH = pKa = 4.74, where each form is half. pH 6.0 is about 1.3 units above the pKa; going up the line at pH 6, the dashed CH₃COO⁻ curve is near 0.95: [A⁻]/[HA] = 10^(6.0 − 4.74) ≈ 18, so A⁻ is 95%. Choosing 5% reads the CH₃COOH curve instead. The rule: when the pH is above the pKa, the base form (A⁻) predominates.
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