Calculate pH of a 0.021 M NaCN Solution
Use this premium cyanide hydrolysis calculator to estimate pH, pOH, hydroxide concentration, and base dissociation behavior for aqueous sodium cyanide. The default setup evaluates a 0.021 M NaCN solution at 25 degrees Celsius using the weak-acid equilibrium of HCN.
How to calculate the pH of a 0.021 M NaCN solution
Sodium cyanide, NaCN, is a salt that dissociates essentially completely in water to produce sodium ions and cyanide ions. The sodium ion is a spectator ion for acid-base chemistry, but the cyanide ion is the conjugate base of hydrocyanic acid, HCN. Because HCN is a weak acid, its conjugate base CN– reacts with water to generate hydroxide, OH–. That makes an aqueous NaCN solution basic. If you are trying to calculate the pH of a 0.021 M NaCN solution, the key idea is that this is not a strong base problem in the same sense as NaOH. Instead, it is a hydrolysis equilibrium problem.
The chemistry starts with two steps. First, NaCN dissolves:
Then the cyanide ion hydrolyzes water:
Since CN– is a weak base, we use the base dissociation constant Kb. In practice, we usually start from the acid dissociation constant of HCN and convert it using:
At 25 degrees Celsius, a commonly used value for Ka of HCN is about 6.2 × 10-10, while Kw = 1.0 × 10-14. That gives:
Set up the ICE table
For a 0.021 M NaCN solution, the initial cyanide concentration is 0.021 M. Let x be the amount of CN– that reacts with water.
- Initial: [CN–] = 0.021, [HCN] = 0, [OH–] = 0
- Change: [CN–] = -x, [HCN] = +x, [OH–] = +x
- Equilibrium: [CN–] = 0.021 – x, [HCN] = x, [OH–] = x
Substitute into the equilibrium expression:
Using Kb ≈ 1.61 × 10-5, the exact quadratic solution gives x ≈ 5.74 × 10-4 M. Since x is the hydroxide concentration:
- [OH–] = 5.74 × 10-4 M
- pOH = -log(5.74 × 10-4) ≈ 3.241
- pH = 14.000 – 3.241 ≈ 10.759
So the pH of a 0.021 M NaCN solution at 25 degrees Celsius is approximately 10.76. This is the value most chemistry instructors and textbook answer keys expect when standard constants are used.
Why NaCN makes water basic
This result often surprises students because NaCN is a salt, and many salts are neutral. The difference is the origin of the ions. Sodium ion comes from the strong base NaOH, so it has negligible acid-base effect in water. Cyanide ion, however, is the conjugate base of a weak acid, HCN. Conjugate bases of weak acids usually produce basic solutions because they remove protons from water and form hydroxide.
You can classify salts in water with a simple rule set:
- Strong acid + strong base salts are usually neutral.
- Strong acid + weak base salts are usually acidic.
- Weak acid + strong base salts are usually basic.
- Weak acid + weak base salts require comparison of Ka and Kb.
NaCN falls into the third category. It comes from the weak acid HCN and the strong base NaOH, so the solution is basic.
Approximation versus exact quadratic solution
For many weak acid and weak base problems, chemistry students use the shortcut x ≈ √(Kb × C). Here, C is the initial concentration of the weak base. For the NaCN problem:
This gives pOH ≈ 3.235 and pH ≈ 10.765. The shortcut answer is nearly the same as the quadratic answer because x is only a small fraction of the initial concentration. Specifically, x / 0.021 is about 2.7%, which satisfies the common 5% rule for weak equilibrium approximations.
Step-by-step expert method you can reuse
1. Identify the acid-base nature of the salt
Ask whether the anion or cation comes from a weak parent acid or base. In NaCN, the anion CN– is the important species.
2. Write the hydrolysis reaction
Always write the base reaction with water:
3. Convert Ka to Kb if needed
If your data table gives Ka of HCN instead of Kb of CN–, use:
4. Create the ICE table
This is the most reliable way to avoid sign mistakes. Start with the analytical concentration of cyanide from sodium cyanide dissolution.
5. Solve for x
Use either the exact quadratic equation or the square-root approximation when justified.
6. Convert to pOH and pH
Because hydroxide is produced directly, pOH often comes first:
Comparison table: pH of NaCN solutions at different concentrations
The table below uses Ka(HCN) = 6.2 × 10-10 and Kw = 1.0 × 10-14 at 25 degrees Celsius. Values are based on the exact equilibrium solution. This gives you context for how strongly concentration affects the pH of sodium cyanide solutions.
| NaCN concentration (M) | Kb of CN– | [OH–] at equilibrium (M) | pOH | pH |
|---|---|---|---|---|
| 0.001 | 1.61 × 10-5 | 1.19 × 10-4 | 3.923 | 10.077 |
| 0.005 | 1.61 × 10-5 | 2.76 × 10-4 | 3.560 | 10.440 |
| 0.010 | 1.61 × 10-5 | 3.93 × 10-4 | 3.406 | 10.594 |
| 0.021 | 1.61 × 10-5 | 5.74 × 10-4 | 3.241 | 10.759 |
| 0.050 | 1.61 × 10-5 | 8.89 × 10-4 | 3.051 | 10.949 |
| 0.100 | 1.61 × 10-5 | 1.26 × 10-3 | 2.900 | 11.100 |
Reference constants and physical data
When solving acid-base equilibrium problems, your final answer depends on the constants you use. Introductory chemistry courses often use rounded textbook values, while advanced references may report slightly different constants due to ionic strength assumptions, temperature, or fitting methodology. The table below shows commonly cited values relevant to this calculation.
| Quantity | Typical value at 25 degrees C | Why it matters | Interpretation |
|---|---|---|---|
| Kw of water | 1.0 × 10-14 | Connects pH and pOH; used to compute Kb from Ka | Higher Kw would slightly shift calculated pH |
| Ka of HCN | 6.2 × 10-10 | Defines the weakness of hydrocyanic acid | Small Ka means HCN is weak, so CN– is a measurable weak base |
| Kb of CN– | 1.61 × 10-5 | Direct equilibrium constant for cyanide hydrolysis | Larger than many weak bases, so NaCN solutions are noticeably basic |
| pKa of HCN | About 9.21 | Useful for buffer and speciation reasoning | Near this pH, HCN and CN– can coexist in significant proportions |
Common mistakes when students calculate the pH of NaCN
Mistake 1: Treating NaCN as a neutral salt
This happens when students look only at the sodium ion and forget that cyanide is the conjugate base of a weak acid. The result is not pH 7. It is clearly basic.
Mistake 2: Using Ka directly in the hydrolysis expression
The hydrolysis reaction of cyanide uses Kb, not Ka. If only Ka is given, convert first.
Mistake 3: Assuming [OH–] equals 0.021 M
That would be true for a strong base with 1:1 stoichiometric release of hydroxide, such as NaOH, not for a weak base like CN–. Only a small fraction of cyanide reacts with water.
Mistake 4: Forgetting to calculate pOH first
Because the ICE table gives [OH–] directly, pOH is the natural first log step. Then convert pOH to pH.
Mistake 5: Ignoring safety context
Although this page is about equilibrium calculation, cyanide compounds are highly hazardous. Real laboratory or industrial handling requires strict safety protocols, proper ventilation, and institutional oversight. This page is educational and computational only.
How concentration changes the final pH
Increasing NaCN concentration increases the amount of cyanide available to react with water, so hydroxide concentration rises and pH increases. However, pH does not scale linearly with concentration because the logarithmic nature of pH and the equilibrium expression both compress the response. For example, going from 0.010 M to 0.100 M NaCN is a tenfold concentration increase, but the pH only rises from about 10.59 to 11.10 with the constants used here. That is a meaningful shift, yet not an enormous one, because weak-base hydrolysis still limits the amount of hydroxide produced.
This is a useful concept in analytical chemistry and environmental chemistry. Weak-base salts can produce moderate but significant alkalinity, and the exact pH depends not only on formal concentration but also on temperature, ionic strength, and the acid-base constants selected from the reference source.
Authoritative references for acid-base constants and cyanide chemistry
- NIST Chemistry WebBook for authoritative thermochemical and chemical reference data.
- NIH PubChem entry for hydrogen cyanide for physical and chemical property summaries.
- LibreTexts Chemistry for educational explanations of salt hydrolysis, conjugate acids and bases, and equilibrium calculations.
Final answer for the default problem
If you use 0.021 M NaCN, Ka(HCN) = 6.2 × 10-10, and Kw = 1.0 × 10-14 at 25 degrees Celsius, then:
- Kb(CN–) ≈ 1.61 × 10-5
- [OH–] ≈ 5.74 × 10-4 M
- pOH ≈ 3.241
- pH ≈ 10.759
This calculator lets you explore how that answer shifts when you change concentration, constants, or the computational method. For most standard chemistry homework, reporting the pH as 10.76 is appropriate.