Calculate the pH of 0.15 M NaCN Solution
Use this premium cyanide hydrolysis calculator to determine pH, pOH, hydroxide concentration, percent hydrolysis, and related equilibrium values for an aqueous sodium cyanide solution. By default, it uses a common literature value for the acid dissociation constant of HCN at 25 degrees Celsius.
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How to calculate the pH of 0.15 M NaCN solution
Sodium cyanide, NaCN, is a salt that fully dissociates in water into Na+ and CN–. The sodium ion is essentially a spectator ion for acid-base chemistry, but the cyanide ion is important because it is the conjugate base of hydrocyanic acid, HCN, a weak acid. That means CN– reacts with water to produce hydroxide ions, OH–, making the resulting solution basic. So when you are asked to calculate the pH of a 0.15 M NaCN solution, you are solving a weak base hydrolysis problem, not a strong base problem.
The key equilibrium is:
CN– + H2O ⇌ HCN + OH–
Because cyanide accepts a proton from water, hydroxide is formed. The more hydroxide produced, the larger the pH. To solve the problem correctly, you first need the base dissociation constant of cyanide, Kb. Most references provide the acid dissociation constant, Ka, for HCN. Once Ka is known, you calculate Kb from the water ion product relation:
Kb = Kw / Ka
Using a typical 25 degrees Celsius value of Ka(HCN) = 6.2 × 10-10 and Kw = 1.0 × 10-14, you get:
Kb = (1.0 × 10-14) / (6.2 × 10-10) = 1.61 × 10-5
Now build the equilibrium setup. If the initial cyanide concentration is 0.15 M, and x is the amount that hydrolyzes, then:
- Initial [CN–] = 0.15 M
- Change = -x for CN–, +x for HCN, +x for OH–
- Equilibrium [CN–] = 0.15 – x
- Equilibrium [HCN] = x
- Equilibrium [OH–] = x
Substitute into the base equilibrium expression:
Kb = [HCN][OH–] / [CN–] = x2 / (0.15 – x)
If you use the common approximation for weak bases, where x is small compared with 0.15, then:
x ≈ √(KbC) = √((1.61 × 10-5)(0.15)) ≈ 1.56 × 10-3 M
Since x equals the hydroxide concentration, then:
[OH–] ≈ 1.56 × 10-3 M
Next compute pOH:
pOH = -log[OH–] ≈ 2.81
And finally:
pH = 14.00 – 2.81 = 11.19
Why NaCN gives a basic solution
This question often appears in general chemistry because it tests whether you can identify the acid-base character of ions from salts. NaCN is formed from a strong base, NaOH, and a weak acid, HCN. Salts of a strong base and weak acid usually produce basic solutions because the anion hydrolyzes water. In contrast, salts made from strong acid and strong base, such as NaCl, are approximately neutral.
A useful way to classify salt behavior is:
- Strong acid + strong base: neutral solution
- Strong acid + weak base: acidic solution
- Weak acid + strong base: basic solution
- Weak acid + weak base: depends on relative Ka and Kb
NaCN clearly falls into the third category. The cyanide ion is the conjugate base of HCN, so it reacts with water and raises the pH.
Exact method versus approximation
For a rigorous answer, especially in educational or lab settings, the exact quadratic approach is better. Starting from:
Kb = x2 / (0.15 – x)
Rearrange into standard quadratic form:
x2 + Kbx – Kb(0.15) = 0
The physically meaningful root is:
x = [-Kb + √(Kb2 + 4KbC)] / 2
Substituting Kb = 1.6129 × 10-5 and C = 0.15 gives:
x ≈ 1.548 × 10-3 M
Then:
- [OH–] = 1.548 × 10-3 M
- pOH = 2.810
- pH = 11.190
Notice that the approximation and exact answer are extremely close. The reason is that only about 1 percent of the cyanide hydrolyzes, so x is indeed small relative to 0.15 M.
| Method | Calculated [OH-] (M) | pOH | pH | Comment |
|---|---|---|---|---|
| Approximation x = √(KbC) | 1.555 × 10-3 | 2.808 | 11.192 | Fast and usually acceptable for homework checks |
| Exact quadratic solution | 1.548 × 10-3 | 2.810 | 11.190 | Preferred when you want the most defensible answer |
How much cyanide actually hydrolyzes
One useful statistic is the percent hydrolysis. This tells you what fraction of the original cyanide concentration reacts with water:
Percent hydrolysis = (x / C) × 100
With x = 1.548 × 10-3 M and C = 0.15 M:
Percent hydrolysis ≈ (1.548 × 10-3 / 0.15) × 100 = 1.03%
That small percentage explains why the approximation works well. It also shows that most cyanide remains as CN– at equilibrium, while a much smaller portion is converted to HCN and OH–.
Comparison data for different literature Ka values
Different textbooks and databases may report slightly different Ka values for HCN depending on source conventions, ionic strength assumptions, or rounding. Those small differences produce small shifts in the final pH. The table below shows how the answer changes for the same 0.15 M NaCN problem when different Ka values are used.
| Ka of HCN | Kb of CN- | Exact [OH-] (M) for 0.15 M NaCN | pH at 25 degrees Celsius |
|---|---|---|---|
| 4.9 × 10-10 | 2.04 × 10-5 | 1.740 × 10-3 | 11.241 |
| 6.2 × 10-10 | 1.61 × 10-5 | 1.548 × 10-3 | 11.190 |
| 6.17 × 10-10 | 1.62 × 10-5 | 1.552 × 10-3 | 11.191 |
This is a good reminder that chemistry answers can vary slightly depending on the equilibrium constants selected. If your teacher or textbook provides a specific Ka or pKa, always use that value.
How concentration affects pH
The pH also depends on the initial NaCN concentration. More cyanide means more available base, so the pH increases as concentration rises. Here is a comparison using Ka(HCN) = 6.2 × 10-10 and the exact solution method.
| NaCN concentration (M) | Exact [OH-] (M) | pOH | pH |
|---|---|---|---|
| 0.010 | 3.935 × 10-4 | 3.405 | 10.595 |
| 0.050 | 8.901 × 10-4 | 3.051 | 10.949 |
| 0.150 | 1.548 × 10-3 | 2.810 | 11.190 |
| 0.500 | 2.831 × 10-3 | 2.548 | 11.452 |
Common mistakes students make
- Treating NaCN as a strong base. NaCN is not the same as NaOH. It forms a basic solution only because CN– hydrolyzes.
- Using Ka directly instead of converting to Kb. For cyanide in water, Kb is the active equilibrium constant.
- Forgetting to calculate pOH first. Since the hydrolysis produces OH–, the direct logarithm step gives pOH, not pH.
- Ignoring temperature effects. If the problem is not at 25 degrees Celsius, Ka and Kw may differ from standard values.
- Using the approximation without checking. The approximation is acceptable only if x is small relative to the starting concentration.
Practical and safety context
Cyanide chemistry matters far beyond the classroom. Cyanide compounds are relevant in mining, electroplating, industrial synthesis, environmental remediation, and toxicology. Because cyanide species are hazardous, real laboratory work involving NaCN or HCN requires strict controls, proper waste handling, and trained supervision. While this page focuses on equilibrium calculations, it is also useful to understand the real-world importance of pH in cyanide systems: lower pH can shift equilibrium toward HCN gas, which is highly dangerous. That is one reason acid-base calculations for cyanide are especially important in environmental and safety management.
Authoritative references for cyanide and acid-base data
For deeper study, consult these high-quality sources:
- U.S. Environmental Protection Agency cyanide resources
- NIST Chemistry WebBook entry for hydrogen cyanide
- University-level overview of acid-base properties of salts
Fast summary
- NaCN dissociates completely to Na+ and CN–.
- CN– is the conjugate base of weak acid HCN, so the solution is basic.
- Compute Kb = Kw / Ka.
- Set up Kb = x2 / (C – x) with C = 0.15 M.
- Solve for x = [OH–].
- Find pOH, then convert to pH.
- For standard constants, the pH is about 11.19.
If you need to calculate the pH of 0.15 M NaCN solution quickly, remember the final number is around 11.19, but the chemistry behind it is what matters: NaCN is basic because cyanide is a weak base in water. The calculator above lets you adjust Ka, Kw, and method so you can match textbook data, lab conditions, or alternative reference values.