An equilibrium lab: FeSCN²⁺
The product of Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq) is blood red, so the depth of color shows the position of equilibrium directly:
- Adding a reactant (Fe³⁺ or SCN⁻) → Q < K → shift right; the color deepens.
- Removing a reactant: Ag⁺ precipitates AgSCN, or OH⁻ precipitates Fe(OH)₃ → Q > K → shift left; the color fades.
- Changing the temperature changes K itself: the forward reaction is exothermic, so heating makes K smaller and the color fades.
- Measuring [FeSCN²⁺] with a spectrophotometer and using an ICE table gives K (4.4.3 + U7.7).
A buffer lab
- Preparing: choose a weak acid with pKa close to the target pH and mix it with its conjugate base in the ratio given by pH = pKa + log([A⁻]/[HA]).
- Measuring pH: calibrate the pH meter with standard buffers of known pH before use.
- Testing: add equal amounts of strong acid or base to the buffer and to pure water and compare the pH changes; once more is added than the buffer components can absorb, the capacity is used up.
An electrochemistry lab: galvanic cells
- Each metal sits in a solution of its own ion; the half-cells are joined by a salt bridge (such as KNO₃) and the wire by a voltmeter.
- The electrode with the larger E° is the cathode (positive, where reduction happens) and the smaller one the anode (negative, where oxidation happens); E°cell = E°cathode − E°anode.
- Electrons flow through the wire from anode to cathode; ions move in the salt bridge to keep each solution electrically neutral.
- The measured voltage is often a little below E°cell: concentrations aren’t exactly 1 M, the temperature isn’t 25 °C, and the wires and solutions have resistance.
Fe³⁺(aq) + SCN⁻(aq) ⇌ FeSCN²⁺(aq). Fe³⁺ is pale yellow, SCN⁻ colorless and FeSCN²⁺ blood red: the deeper the color, the more FeSCN²⁺. The forward reaction is exothermic.
On the left is 50.0 mL of pure water; on the right, 50.0 mL of a buffer (0.100 M CH₃COOH and 0.100 M CH₃COONa). Add the same amount of 0.100 M HCl or NaOH to both:
Each metal sits in a 1.0 M solution of its own ion; the half-cells are joined by a KNO₃ salt bridge and a voltmeter. Choose two metals:
Example 1: K for FeSCN²⁺ by spectrophotometry
5.00 mL of 2.00 × 10⁻³ M Fe(NO₃)₃ is mixed with 5.00 mL of 2.00 × 10⁻³ M KSCN. At equilibrium, a calibration curve gives [FeSCN²⁺] = 1.50 × 10⁻⁴ M. Find K.
-
Starting concentrations after mixing. The volume doubles, so each concentration halves: [Fe³⁺]₀ = [SCN⁻]₀ = 1.00 × 10⁻³ M.
-
ICE table. 1.50 × 10⁻⁴ M of FeSCN²⁺ forms, using up the same amount of Fe³⁺ and of SCN⁻:
-
Substitute into the K expression.
Example 2: a Ni–Ag cell
Ni dips into 1.0 M Ni²⁺ and Ag into 1.0 M Ag⁺, joined by a salt bridge (E°: Ag⁺/Ag = +0.80 V, Ni²⁺/Ni = −0.26 V). Identify the cathode and anode, write the overall reaction, and find E°cell. How does the voltage change if the Ag⁺ solution is diluted?
-
Cathode and anode. Ag⁺/Ag has the larger E°, so Ag is the cathode (Ag⁺ is reduced) and Ni is the anode (Ni is oxidized).
-
Overall reaction. Balance the electrons (Ag⁺ gains 1, Ni loses 2):
-
E°cell.
-
Diluting Ag⁺. Q = [Ni²⁺]/[Ag⁺]² gets larger, the drive to the right weakens, and the voltage drops (below 1.06 V).
- A few drops of Fe(NO₃)₃ solution are added
More reactant, so Q < K.
- A few drops of KSCN solution are added
More reactant, so Q < K.
- A few drops of AgNO₃ solution are added
Ag⁺ precipitates AgSCN, removing SCN⁻.
- A few drops of NaOH solution are added
OH⁻ precipitates Fe(OH)₃, removing Fe³⁺.
- The test tube is placed in hot water
The forward reaction is exothermic, so heating makes K smaller.
- The test tube is placed in ice water
Cooling makes K larger.
- A little solid KNO₃ is added
K⁺ and NO₃⁻ don’t take part, so Q is unchanged.
- A catalyst is added
A catalyst speeds up the forward and reverse reactions equally and doesn’t change the equilibrium.
Common mistakes
Mistake 1: “Adding any ion shifts the equilibrium.” Only a change that alters the concentration of something in Q does. Spectator ions such as K⁺ and NO₃⁻ leave Q unchanged, so there’s no shift.
Mistake 2: “A buffer can absorb any amount of acid or base.” Its capacity is limited. Once the strong acid added exceeds the A⁻ present (or strong base exceeds the HA), the pH changes sharply.
Mistake 3: “Electrons flow through the salt bridge.” Electrons travel only through the wire; the salt bridge carries moving ions, which keep both solutions electrically neutral and complete the circuit.
Quick check
What does the salt bridge do in a galvanic cell?
Without it, charge would build up and the reaction would quickly stop.
In a galvanic cell, which way do electrons flow through the wire?
Oxidation at the anode releases electrons, which flow to the cathode.
What should be done before using a pH meter?
Only a calibrated meter reads accurately.
Which weak acid is best for making a buffer with pH ≈ 4.7?
The closer the pKa to the target pH, the better the buffer.
A little HCl is added to a CH₃COOH/CH₃COO⁻ buffer and the pH barely changes because:
The added H⁺ is used up by the conjugate base, so [A⁻]/[HA] changes only slightly.
A Cu–Zn cell has E°cell = 1.10 V but reads 1.06 V. The most reasonable explanation is:
Nonstandard conditions and resistance both make the measured voltage a little lower.
AP-style practice
A student is to prepare a buffer with pH 5.00 from 0.200 M CH₃COOH and solid CH₃COONa (82.03 g/mol). For CH₃COOH, Ka = 1.8 × 10⁻⁵ (pKa = 4.74).
(a) Why is acetic acid a good choice for this buffer?
(b) Calculate the mass of CH₃COONa to add to 100.0 mL of 0.200 M CH₃COOH (ignore any volume change).
(c) Briefly describe how to prepare the buffer and check its pH, naming the equipment.
(d) After preparation, the pH meter reads 5.12. Suggest one possible reason.
(e) 1.00 mL of 1.00 M HCl is added to the buffer. Give the net ionic equation for what happens, and calculate the new pH.
Model answer and scoring
(a) Acetic acid’s pKa of 4.74 is close to the target pH of 5.00, so [A⁻]/[HA] is close to 1 and the buffer works well.
(b) 5.00 = 4.74 + log([A⁻]/[HA]), so [A⁻]/[HA] = 10^0.26 = 1.82; [A⁻] = 1.82 × 0.200 = 0.364 M; n = 0.364 × 0.1000 = 0.0364 mol; m = 0.0364 × 82.03 = 2.99 g.
(c) Measure 100.0 mL of 0.200 M CH₃COOH accurately (with a pipet or volumetric flask) into a beaker; weigh 2.99 g of CH₃COONa on an analytical balance and stir it in until it dissolves; calibrate a pH meter with standard buffers, then measure the pH.
(d) For example: the pH meter wasn’t calibrated; too much CH₃COONa was weighed out ([A⁻]/[HA] too large, so the pH is too high); the CH₃COOH solution is actually weaker than 0.200 M.
(e) CH₃COO⁻(aq) + H⁺(aq) → CH₃COOH(aq). Adding 1.00 × 10⁻³ mol of H⁺ gives n(A⁻) = 0.0364 − 0.0010 = 0.0354 mol and n(HA) = 0.0200 + 0.0010 = 0.0210 mol; pH = 4.74 + log(0.0354 ÷ 0.0210) = 4.97.
Scoring: (a) pKa close to the target pH; (b) the ratio, moles and mass; (c) a sensible procedure and equipment, including calibrating the pH meter; (d) one reasonable cause of a pH that is too high; (e) the net ionic equation and the new pH.
Summary
- FeSCN²⁺ lab: the color shows the position of equilibrium; adding or removing a species in Q shifts it, and changing the temperature changes K; with spectrophotometry and an ICE table you can find K.
- Buffers: choose a weak acid with pKa near the target pH and use the Henderson-Hasselbalch equation for the ratio; calibrate the pH meter first; buffer capacity is limited.
- Galvanic cells: the electrode with the larger E° is the cathode; electrons flow through the wire from anode to cathode while ions move in the salt bridge; a measured voltage slightly below E°cell is normal.
AP tip: When a lab question asks what is “observed,” describe something you can see (the color deepens, a precipitate forms, the voltmeter reading), then follow with the explanation in terms of Q and K.
Sources: Ka of CH₃COOH from OpenStax Chemistry 2e Appendix H; standard reduction potentials from Appendix L of the same book; the data for the Fe³⁺–SCN⁻ lab are illustrative.