Chemistry Practice Problems
Generated chemistry practice problems on concentration, buffers, titration, half-life and yield, marked to within 1.5 percent with the working shown.
Practice
Question 1 of 40
Henderson-Hasselbalch Calculator
- pH
- pH = 3.2
- Conjugate base [A⁻]
- [A⁻] = 86 mM
- Weak acid [HA]
- [HA] = 120 mM
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Worked answers
The first ten questions from the set above, each with its answer and the working that gets there. The working is carried out in the units each equation takes, so its last line can show the answer before it is converted.
-
- pH
- 3.2
- Conjugate base [A⁻]
- 86 mM
- Weak acid [HA]
- 120 mM
Find the pKa of the acid.
Show the answer and working
Answer pKa = 3.345
Rearranged
pKa = pH − log₁₀([A⁻] / [HA])- pKa = pH - log10([A-] / [HA])
- = 3.2 - log10(0.71667)
- = 3.2 + 0.1447
- = 3.345
Check it with the Henderson-Hasselbalch Calculator.
-
- Mass of solute
- m = 14 g
- Final volume
- V = 1300 mL
Find the concentration (%).
Show the answer and working
Answer Concentration = 1.077 % w/v
Rearranged
% w/v = m ÷ V × 100, with m in g and V in mL- % w/v = mass / volume x 100, mass in g and volume in mL
- = 14 / 1300 x 100
- = 0.010769 x 100
- = 1.077 % w/v
Check it with the Percent Solution Calculator.
-
- Starting amount
- N₀ = 170
- Half-life
- t½ = 7300 years
- Time elapsed
- t = 23,000 years
Find the amount remaining (N).
Show the answer and working
Answer N = 19.14
Rearranged
N = N₀ × (½)^(t / t½)- N = N0 x (1/2)^n, where n = t / t_half
- n = 3.151 half-lives elapsed
- = 170 x (1/2)^3.151
- = 19.142
Check it with the Half-Life Calculator.
-
- Analyte concentration
- c₁ = 86 mM
- Analyte volume
- V₁ = 96 mL
- Titrant volume at endpoint
- V₂ = 50 mL
- Mole ratio, titrant per analyte
- r = 1
Find the titrant concentration (c₂).
Show the answer and working
Answer c₂ = 165.1 mM
Rearranged
c₂ = r · c₁V₁ ÷ V₂- r x c1 x V1 = c2 x V2 at the endpoint, with r = 1
- c2 = r c1 V1 / V2
- = 1 x 86 mM x 96 mL / 50 mL
- = 165.12 mM = 0.16512 mol/L
Check it with the Titration Calculator.
-
- Theoretical yield
- 9.3 g
- Percent yield
- 95 %
Find the actual yield.
Show the answer and working
Answer Actual yield = 8.835 g
Rearranged
actual = theoretical × % ÷ 100- actual = theoretical x % / 100
- = 9.3 x 95 / 100
- = 883.5 / 100
- = 8.835 g
Check it with the Percent Yield Calculator.
-
- Rate constant at T₁
- k₁ = 0.028
- Second temperature
- T₂ = 320 K
- Rate constant at T₂
- k₂ = 0.12
- Activation energy
- Ea = 16 kJ/mol
Find the first temperature (T₁).
Show the answer and working
Answer T₁ = 257.6 K
Rearranged
1/T₁ = 1/T₂ + (R/Ea) ln(k₂/k₁)- ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)
- 1/T1 = 1/T2 + (R/Ea) ln(k2/k1)
- = 1/320 + (8.31446 / 16,000) x ln(0.12 / 0.028)
- = 0.003125 + 0.00075625 = 0.0038812 per K
- T1 = 257.65 K = -15.5 °C
Check it with the Arrhenius Equation Calculator.
-
- pH
- 4.2
- pKa of the acid
- 4.3
- Conjugate base [A⁻]
- 180 mM
Find the weak acid [HA].
Show the answer and working
Answer [HA] = 226.6 mM
Rearranged
[HA] = [A⁻] / 10^(pH − pKa)- [HA] = [A-] / 10^(pH - pKa)
- = 180 / 10^(4.2 - 4.3)
- = 180 / 10^-0.1
- = 226.61 mM
Check it with the Henderson-Hasselbalch Calculator.
-
- Mass of solute
- m = 4.8 g
- Final volume
- V = 1600 mL
Find the concentration (%).
Show the answer and working
Answer Concentration = 0.3 % w/v
Rearranged
% w/v = m ÷ V × 100, with m in g and V in mL- % w/v = mass / volume x 100, mass in g and volume in mL
- = 4.8 / 1600 x 100
- = 0.003 x 100
- = 0.3 % w/v
Check it with the Percent Solution Calculator.
-
- Starting amount
- N₀ = 350
- Amount remaining
- N = 29
- Half-life
- t½ = 9000 years
Find the time elapsed (t).
Show the answer and working
Answer t = 32,340 years
Rearranged
t = t½ · log₂(N₀ / N)- t = t_half x log2(N0 / N)
- N0 / N = 12.069
- log2(12.069) = 3.593 half-lives
- t = 3.593 x 9 thousand yr = 32.339 thousand yr
Check it with the Half-Life Calculator.
-
- Analyte volume
- V₁ = 62 mL
- Titrant concentration
- c₂ = 170 mM
- Titrant volume at endpoint
- V₂ = 56 mL
- Mole ratio, titrant per analyte
- r = 1
Find the analyte concentration (c₁).
Show the answer and working
Answer c₁ = 153.5 mM
Rearranged
c₁ = c₂V₂ ÷ (r · V₁)- r x c1 x V1 = c2 x V2 at the endpoint, with r = 1
- c1 = c2 V2 / (r V1)
- = (170 mM x 56 mL) / (1 x 62 mL)
- = 153.55 mM = 0.15355 mol/L
Check it with the Titration Calculator.
Everything goes through moles and litres
Put each quantity into the unit the formula wants before rearranging. Concentrations
arrive in millimolar and micromolar where the formulas want moles per litre, so
15 µM is 1.5 × 10⁻⁵ M, and volumes arrive in millilitres.
Path length is the exception: extinction coefficients are quoted per centimetre and a
standard cuvette is 1 cm.
Anything starting from a mass needs a molar mass, and that depends on the hydration state: anhydrous copper sulfate is 159.60 g/mol against 249.68 for the pentahydrate, so treating one as the other leaves you about 36 percent short. A reaction needs the balanced equation on paper, because the mole ratio is the number most likely to be assumed rather than read.
The errors that survive a sanity check
- A prefix carried into the formula. 15 µM in a 1 cm cuvette with an
extinction coefficient of 43,000 M⁻¹cm⁻¹ gives an absorbance of 0.645. Substituting
15 rather than
1.5 × 10⁻⁵returns 645,000, which no spectrophotometer could report. - A mole ratio assumed to be one. Sulfuric acid supplies two protons, so 2.5 mmol of it needs 50 mL of 100 mM sodium hydroxide, not 25 mL. An answer wrong by exactly two is almost never arithmetic.
- The buffer ratio inverted. pH is the pKa plus the log of base over acid. With 60.8 mM conjugate base against 39.2 mM acid and a pKa of 7.21 the answer is 7.40; flipping the ratio returns 7.02, which still looks like a pH.
- Percent without saying which percent. Weight per volume, weight per weight and volume per volume are all written as percent and can differ by a factor of two for the same number.
Checks that take one line
Half-life problems are the easiest to verify, because a whole number of half-lives has a fixed answer: the fraction left is one over two to that power. Falling from 100 to 25 is exactly two half-lives, whatever the isotope, so 11,400 years of carbon-14 decay at a half-life of 5700 years needs no calculation at all.
Buffers offer the same check: equal concentrations of acid and base put the pH exactly at the pKa, and a ten to one ratio moves it one unit. 1 percent w/v is 10 g/L, so 0.9 percent saline is 9 g/L and, divided by the 58.44 g/mol of sodium chloride, 154 mM. A percent yield above 100 is not a good result, it is a wet or impure product.
What a concentration unit actually claims
Molarity is moles per litre of finished solution, so it shifts as the solution expands with temperature. Molality is moles per kilogram of solvent, and does not. In dilute aqueous solution 1 mg/L is taken as 1 ppm, and 1 percent w/v is also 10 mg/mL, which is often the more useful form.
Two conventions sit underneath the pH numbers. pH plus pOH comes to 14 only at 25 °C, because that sum is pKw and pKw depends on temperature. A pKa is itself temperature dependent: Tris shifts by roughly 0.03 pH units per degree Celsius, so a buffer set to pH 8.0 on the bench is nearer 8.5 in a cold room.
Common questions
Are the answers checked, or just computed?
They are computed by the same solver that runs the corresponding calculator on this site, so there is no separate answer key that could drift out of step with the code. That is the whole reason the questions are generated rather than written: a stated answer and the calculator it came from cannot disagree. Those solvers are covered by the test suite, and each question shows the substitution it used so you can see where a value entered the formula.
Do the same questions come back on my next visit?
Yes. The set is a fixed pool of questions generated when the site is built, the same for everyone, and each visit starts at a random point in it. Nothing is stored between visits: there is no account, no saved progress and no record of what you answered last time. Within a single visit you get a running count of how many you have got right out of how many you have attempted, and that is the only tally kept.