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B.2 Algebra

Syllabus
2017
Topic
—
Level
A2

Learning objectives

—B.2.1—Mathematical symbolsUnderstand and use =, <, ≪, ≫, >, ∝, ~ and the equilibrium sign.—B.2.2—Changing the subject of an equationChange the subject of an equation in structured and unstructured chemical calculations. A2 applications include calculating a rate constant from a rate equation.—B.2.3—Substitution into algebraic equationsSubstitute numerical values into algebraic equations using appropriate physical units, including mole calculations. A2 applications include rate and equilibrium-constant calculations.—B.2.4—Solving algebraic equationsSolve algebraic equations, including Hess's law calculations. A2 applications include calculating a rate constant from a rate equation.—B.2.5—Logarithmic quantitiesUse logarithms for quantities spanning several orders of magnitude, including pH and pKa calculations.

Read every mathematical symbol as a precise chemical claim

Symbol Meaning in a calculation or chemical statement
== both sides have the same value
<<, >> strictly less than, strictly greater than
≪\ll, ≫\gg much smaller than, much greater than on the relevant scale
∝\propto proportional: one quantity equals a constant times the other
∼\sim an approximate relation; its exact sense must come from context
⇌\rightleftharpoons forward and reverse reactions occur and can establish dynamic equilibrium

If rate ∝[A]\propto[\ce{A}], then rate =k[A]=k[\ce{A}] for a fixed set of conditions. The proportionality sign does not mean the numerical values are equal: the constant kk supplies the scale and units.

At dynamic equilibrium, the forward and reverse rates are equal, so macroscopic concentrations remain constant. The equilibrium sign does not mean equal concentrations, complete reaction or that particles have stopped reacting.

Symbols are not decorative shorthand. Replace ∝\propto by == only after introducing a proportionality constant, and use ≪\ll or ≫\gg only when the relative scale makes 'much smaller' or 'much larger' defensible.

Rearrange the relationship before inserting numbers

Structure To isolate the target
target multiplied by a factor divide both sides by that factor
target divided by a factor multiply both sides by that factor
target raised to a power apply the matching root
target inside several factors preserve brackets, then undo operations in reverse order

\text{rate}=k[\ce{A}]^2[\ce{B}]\quad\Longrightarrow\quad k=\frac{\text{rate}}{[\ce{A}]^2[\ce{B}]}

If rate is 0.040 mol dm−3 s−10.040\,\mathrm{mol\,dm^{-3}\,s^{-1}}, [A]=0.010 mol dm−3[\ce{A}]=0.010\,\mathrm{mol\,dm^{-3}} and [B]=0.050 mol dm−3[\ce{B}]=0.050\,\mathrm{mol\,dm^{-3}}, then k=8.0imes103 dm6 mol−2 s−1k=8.0 imes10^3\,\mathrm{dm^6\,mol^{-2}\,s^{-1}}. The unit follows by dividing the rate unit by three concentration factors.

Check the rearrangement symbolically before substitution: multiply the final expression back by the removed factor and confirm that the original equation returns. This separates an algebra error from a calculator-entry error.

An operation applied to one side must be applied to the entire other side. Do not cancel a term across addition or subtraction, and do not lose an exponent when moving a concentration factor.

Substitution is a traceable chain of values and units

Step Action
define write the equation and identify each symbol
align convert measurements to the units required by the equation
substitute place each value, unit and power in the correct position
calculate keep guard digits and preserve brackets
report attach the derived unit and justified significant figures

n=cV

For c=0.200 mol dm−3c=0.200\,\mathrm{mol\,dm^{-3}} and V=25.0 cm3=0.0250 dm3V=25.0\,\mathrm{cm^3}=0.0250\,\mathrm{dm^3}, n=0.200imes0.0250=5.00imes10−3 moln=0.200 imes0.0250=5.00 imes10^{-3}\,\mathrm{mol}. The conversion is part of the substitution, not an optional correction after calculating.

At A2, substitute equilibrium concentrations into the stated KcK_c expression and preserve every stoichiometric power; for rates, distinguish rate from rate constant and derive the unit of kk from the rate equation. A multi-stage calculation should label intermediate quantities so each value can be traced to its source.

Never substitute a raw volume in cm3\mathrm{cm^3} into an equation expecting dm3\mathrm{dm^3}, or omit brackets around a negative value or powered concentration. A correct-looking number without the required unit is not a complete physical result.

Build one algebraic equation from the chemical route

Solving a chemical equation begins by translating the route, conservation rule or definition into one algebraic statement. Hess's law works because enthalpy is a state function: the total enthalpy change between the same initial and final states is independent of the route.

Change to a reaction step Change to ΔH\Delta H
reverse the equation reverse the sign
multiply every coefficient by nn multiply ΔH\Delta H by nn
add reaction equations add their adjusted ΔH\Delta H values

\Delta H_{A\to C}=\Delta H_{A\to B}+\Delta H_{B\to C}

If ΔHAoC=−120 kJ mol−1\Delta H_{A o C}=-120\,\mathrm{kJ\,mol^{-1}} and ΔHAoB=−50 kJ mol−1\Delta H_{A o B}=-50\,\mathrm{kJ\,mol^{-1}}, then −120=−50+x-120=-50+x, so x=−70 kJ mol−1x=-70\,\mathrm{kJ\,mol^{-1}}. Substitution back gives −50+(−70)=−120-50+(-70)=-120, confirming both magnitude and sign.

The same discipline applies to an unknown in a rate equation: construct the correct relationship, isolate the unknown, solve, then test the result in the original equation with units.

Do not change an enthalpy sign merely because a value moves across an equals sign; the sign changes when the chemical step is reversed. Coefficients and enthalpy must be scaled together.

Logarithms turn orders of magnitude into an additive scale

A base-10 logarithm reports the power to which 10 must be raised. This compresses concentrations spanning many powers of ten: changing a value by a factor of 10 changes its logarithmic measure by one unit.

\mathrm{pH}=-\log_{10}[\ce{H+}]\qquad \mathrm{p}K_a=-\log_{10}K_a

Given Recover
[HX+][\ce{H+}] pH with −log⁡10-\log_{10}
pH [HX+]=10−pH[\ce{H+}]=10^{-\mathrm{pH}}
KaK_a pKa with −log⁡10-\log_{10}
pKa Ka=10−pKaK_a=10^{-\mathrm{p}K_a}

If [HX+]=3.2imes10−4 mol dm−3[\ce{H+}]=3.2 imes10^{-4}\,\mathrm{mol\,dm^{-3}}, then pH=3.49\mathrm{pH}=3.49. If Ka=1.8imes10−5K_a=1.8 imes10^{-5}, then pKa=4.74\mathrm{p}K_a=4.74. Substituting each result into its inverse power relation checks the calculator entry.

A lower pH corresponds to a higher hydrogen-ion concentration: a decrease of one pH unit means a tenfold increase in [HX+][\ce{H+}]. Likewise, a lower pKa corresponds to a larger KaK_a.

pH and pKa are full-IAL applications in this specification. They are logarithmic quantities without concentration units; do not omit the minus sign or apply the logarithm to only part of a concentration written in standard form.