Calculate the pH of 0.20 M NaCN
Use this interactive chemistry calculator to determine the pH of a sodium cyanide solution by modeling cyanide as the conjugate base of hydrocyanic acid and solving the weak-base equilibrium accurately.
NaCN pH Calculator
How to calculate the pH of 0.20 M NaCN
To calculate the pH of 0.20 M NaCN, you need to recognize what sodium cyanide does in water. NaCN is a soluble ionic compound that dissociates essentially completely into Na+ and CN–. The sodium ion is a spectator ion for acid-base chemistry, but the cyanide ion is important because it is the conjugate base of hydrocyanic acid, HCN. Since HCN is a weak acid, CN– behaves as a weak base in water.
The key equilibrium is:
CN– + H2O ⇌ HCN + OH–
This means a sodium cyanide solution is basic, not neutral. The pH must therefore be greater than 7 at 25 degrees C. The problem is solved by converting the acid dissociation constant for HCN into a base dissociation constant for CN–, then solving for hydroxide concentration.
Step 1: Write the dissociation and hydrolysis reactions
First, sodium cyanide dissociates completely:
- NaCN(aq) → Na+(aq) + CN–(aq)
The cyanide then hydrolyzes water:
- CN– + H2O ⇌ HCN + OH–
Because NaCN is a strong electrolyte, the initial cyanide concentration is the same as the formal salt concentration. For a 0.20 M solution, the initial concentration of CN– is 0.20 M.
Step 2: Use the relationship between Ka and Kb
For a conjugate acid-base pair at 25 degrees C:
Ka × Kb = Kw = 1.0 × 10-14
Hydrocyanic acid has a commonly used pKa near 9.21 at 25 degrees C. Converting that to Ka:
Ka = 10-9.21 ≈ 6.17 × 10-10
Then the base constant for cyanide is:
Kb = Kw / Ka = (1.0 × 10-14) / (6.17 × 10-10) ≈ 1.62 × 10-5
Step 3: Set up the ICE table
For the reaction CN– + H2O ⇌ HCN + OH–, the ICE table is:
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| CN– | 0.20 | -x | 0.20 – x |
| HCN | 0 | +x | x |
| OH– | 0 | +x | x |
The equilibrium expression is:
Kb = [HCN][OH–] / [CN–] = x2 / (0.20 – x)
Step 4: Solve for x
Substitute the values:
1.62 × 10-5 = x2 / (0.20 – x)
Since Kb is small relative to the starting concentration, you can first use the weak-base approximation:
x ≈ √(KbC) = √((1.62 × 10-5)(0.20))
x ≈ √(3.24 × 10-6) ≈ 1.80 × 10-3 M
This x value is the hydroxide concentration:
[OH–] ≈ 1.80 × 10-3 M
The 5 percent check confirms the approximation is valid:
(1.80 × 10-3 / 0.20) × 100 ≈ 0.90%
Because the change is less than 5 percent, the square-root approximation is acceptable. If you prefer, the exact quadratic solution gives essentially the same answer for this concentration.
Step 5: Convert [OH–] to pOH and pH
Now calculate pOH:
pOH = -log(1.80 × 10-3) ≈ 2.74
Finally:
pH = 14.00 – 2.74 = 11.26
Why NaCN produces a basic solution
Many students first see sodium salts and assume they are neutral because sodium itself comes from a strong base, NaOH. That is only half the story. The anion matters too. Chloride from HCl is neutral in water because HCl is a strong acid, so Cl– has negligible basicity. Cyanide is different because HCN is a weak acid. Since its conjugate base is not negligible, CN– removes protons from water and forms hydroxide. That is the direct chemical reason NaCN solutions are basic.
Common mistakes when solving this problem
- Treating NaCN as a neutral salt. It is not. The CN– ion is basic.
- Using Ka directly instead of converting to Kb. Because cyanide is acting as a base, the hydrolysis equilibrium should use Kb.
- Using 0.20 M as [OH–]. The hydroxide concentration is not the same as the salt concentration. Only a small fraction hydrolyzes.
- Forgetting to calculate pOH first. Weak bases produce OH–, so pOH comes before pH.
- Ignoring the temperature dependence of Kw. In most classroom problems, 25 degrees C is implied and Kw = 1.0 × 10-14.
Exact versus approximate solution
For a weak-base equilibrium, the shortcut x ≈ √(KbC) is often accurate enough. Still, it is useful to know the exact form. Starting from:
Kb = x2 / (C – x)
Rearrange to:
x2 + Kbx – KbC = 0
The positive quadratic root gives the exact [OH–]. For 0.20 M NaCN with Kb ≈ 1.62 × 10-5, the exact solution is essentially 1.79 × 10-3 M, yielding a pH near 11.25. This near-perfect agreement shows why the approximation is standard in chemistry courses.
Comparison table: key constants used in the calculation
| Quantity | Typical value at 25 degrees C | Why it matters |
|---|---|---|
| Formal NaCN concentration | 0.20 M | Sets the initial CN– concentration |
| pKa of HCN | 9.21 | Used to compute Ka for hydrocyanic acid |
| Ka of HCN | 6.17 × 10-10 | Converted into Kb for cyanide |
| Kw | 1.00 × 10-14 | Relates Ka and Kb |
| Kb of CN– | 1.62 × 10-5 | Directly controls OH– formation |
| Calculated pH | 11.25 to 11.26 | Final solution pH for 0.20 M NaCN |
How concentration changes the pH of NaCN solutions
Because cyanide is a weak base, the pH rises as NaCN concentration increases, but not in a perfectly linear way. The governing relationship is tied to the square root of concentration in the common approximation. That means a 100 times increase in concentration does not increase pH by 100 times. Instead, it shifts the hydroxide concentration according to weak-equilibrium behavior.
| NaCN concentration (M) | Approx. [OH–] (M) | Approx. pOH | Approx. pH |
|---|---|---|---|
| 0.001 | 1.27 × 10-4 | 3.90 | 10.10 |
| 0.010 | 4.02 × 10-4 | 3.40 | 10.60 |
| 0.050 | 9.01 × 10-4 | 3.05 | 10.95 |
| 0.20 | 1.80 × 10-3 | 2.74 | 11.26 |
| 0.50 | 2.85 × 10-3 | 2.55 | 11.45 |
| 1.00 | 4.02 × 10-3 | 2.40 | 11.60 |
Interpretation of the answer
A pH near 11.25 means the solution is moderately basic. It is nowhere near as basic as a 0.20 M strong base such as NaOH, but it is clearly alkaline because cyanide undergoes measurable hydrolysis. This distinction between strong and weak bases is central in acid-base chemistry. In a strong base solution, the hydroxide concentration is essentially equal to the analytical concentration. In a weak base solution like cyanide, equilibrium limits the amount of OH– formed.
Real-world and laboratory relevance
Questions involving NaCN are common in general chemistry because they test whether you can classify salts correctly and use conjugate acid-base relationships. Beyond the classroom, cyanide chemistry matters in analytical chemistry, metallurgy, environmental chemistry, and toxicology. Because cyanide compounds are hazardous, pH can strongly influence the fraction present as CN– versus HCN. At lower pH, more protonated HCN can exist, and that has major safety implications. In educational settings, however, the problem is normally focused on equilibrium math rather than handling guidance.
Best method to remember for exams
- Identify whether the salt comes from a weak acid or weak base.
- If the anion is the conjugate base of a weak acid, expect a basic solution.
- Use Kb for the hydrolysis reaction.
- Apply an ICE table.
- Find [OH–], then pOH, then pH.
- Check whether the approximation is valid.
Authoritative references for constants and acid-base background
For students who want to verify acid-base constants or review equilibrium foundations, these sources are useful:
Bottom line
When asked to calculate the pH of 0.20 M NaCN, the correct chemistry is to treat cyanide as a weak base. Using pKa(HCN) ≈ 9.21 gives Kb(CN–) ≈ 1.62 × 10-5. Solving the equilibrium produces [OH–] ≈ 1.8 × 10-3 M, pOH ≈ 2.74, and pH ≈ 11.26. If your answer is near 11.25, you are on the right track.