Units, standard form and proportion in chemical calculation: checking answers, buffers, rate constants and units

Overview

Checking answers by order of magnitude, pKa and buffer pH calculations with their approximations, rate constants and the Arrhenius temperature, units for heat capacity and Kc, and standard form and significant figures.

What you need to know

Answer each one before you reveal it.

Questions from this subtopic

Each one with the model answer underneath.

1. 1

   A student calculates a percentage yield of 112%. What does this tell you, and what are two likely causes?

   Model answer

   A yield above 100% is impossible, because the theoretical mass is the maximum the equation allows. Likely causes are that the product was not fully dried, so residual solvent added to its mass, or that the theoretical mass was calculated wrongly, for example by using the wrong reacting ratio or an incorrect molar mass.

2. 2

   How can an order-of-magnitude check reveal an error in a calculation of the number of molecules in a sample?

   Model answer

   Any laboratory-scale sample contains a number of molecules in the region of 10²⁰ to 10²⁴. If a calculation returns a value such as 4 × 10⁻²² molecules, the Avogadro constant has almost certainly been divided by instead of multiplied. Comparing the exponent with the expected magnitude catches this immediately.

3. 3

   Calculate the pKa of ethanoic acid, given Ka = 1.74 × 10⁻⁵ mol dm⁻³.

   Model answer

   pKa = −log₁₀(Ka) = −log₁₀(1.74 × 10⁻⁵) = 4.76. The negative logarithm converts an inconveniently small number into a value near 5, which is easier to compare between acids.

4. 4

   The pKa of methanoic acid is 3.75. Calculate its Ka, giving the units.

   Model answer

   Ka = 10⁻ᵖᴷᵃ = 10⁻³·⁷⁵ = 1.78 × 10⁻⁴ mol dm⁻³.

5. 5

   Calculate the pH of 0.0200 mol dm⁻³ sodium hydroxide solution at 298 K. (Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶)

   Model answer

   Sodium hydroxide is fully dissociated, so [OH⁻] = 0.0200 mol dm⁻³. [H⁺] = Kw ÷ [OH⁻] = (1.00 × 10⁻¹⁴) ÷ (2.00 × 10⁻²) = 5.00 × 10⁻¹³ mol dm⁻³. pH = −log₁₀(5.00 × 10⁻¹³) = 12.30.

6. 6

   A buffer of pH 5.00 is to be made from ethanoic acid, Ka = 1.74 × 10⁻⁵ mol dm⁻³, and sodium ethanoate. Calculate the ratio [CH₃COO⁻] ÷ [CH₃COOH] required, giving your answer to two significant figures.

   Model answer

   At pH 5.00 the hydrogen ion concentration is [H⁺] = 10⁻⁵·⁰⁰ = 1.00 × 10⁻⁵ mol dm⁻³. Rearranging Ka = [H⁺] × [CH₃COO⁻] ÷ [CH₃COOH] gives [CH₃COO⁻] ÷ [CH₃COOH] = Ka ÷ [H⁺] = (1.74 × 10⁻⁵) ÷ (1.00 × 10⁻⁵) = 1.74, which is 1.7 to two significant figures. The salt must therefore be about 1.7 times as concentrated as the acid.

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