Calculate the pH of a 0.21 M NaCN Solution
This premium calculator solves the pH of sodium cyanide solutions by modeling cyanide as the conjugate base of hydrocyanic acid. Enter or confirm the standard values below, choose your preferred solution method, and instantly see pH, pOH, hydroxide concentration, and a concentration-to-pH chart.
Expert Guide: How to Calculate the pH of a 0.21 M NaCN Solution
If you need to calculate the pH of a 0.21 M NaCN solution, the key idea is that sodium cyanide is not itself an acid. It is a salt made from a strong base, NaOH, and a weak acid, HCN. Because the sodium ion is a spectator ion in water, the chemistry that matters comes from the cyanide ion, CN–. Cyanide behaves as a weak base and reacts with water to produce hydroxide ions. Those hydroxide ions make the solution basic, so the pH ends up above 7.
This is a classic weak-base hydrolysis problem in general chemistry. To solve it properly, you begin by identifying the acid-base relationship, converting the acid dissociation constant of HCN into the base dissociation constant of CN–, and then solving for the equilibrium hydroxide concentration. Once you have [OH–], you find pOH and finally pH. For the standard data set at 25 C using Ka(HCN) = 4.9 × 10-10, the pH of a 0.21 M NaCN solution is approximately 11.31.
Why NaCN makes water basic
Sodium cyanide dissociates completely in water:
NaCN → Na+ + CN–
The sodium ion does not appreciably affect pH, but the cyanide ion does. Because CN– is the conjugate base of the weak acid HCN, it reacts with water:
CN– + H2O ⇌ HCN + OH–
This reaction generates OH–, so the solution is basic. The stronger the basicity of CN–, the more hydroxide is produced. Since HCN is a weak acid, its conjugate base CN– is significantly basic compared with conjugate bases of strong acids, which are essentially neutral.
Step 1: Write the given concentration
The problem states the NaCN concentration is 0.21 M. Because NaCN is a soluble ionic compound, it dissociates almost completely, so the initial cyanide concentration is also taken as:
[CN–]initial = 0.21 M
This becomes the starting concentration in the hydrolysis equilibrium.
Step 2: Convert Ka of HCN into Kb of CN–
You usually know the acid constant for HCN rather than the base constant for CN–. At 25 C, the relationship between conjugate acid-base pairs is:
Ka × Kb = Kw
Using standard values:
- Ka(HCN) = 4.9 × 10-10
- Kw = 1.0 × 10-14 at 25 C
So:
Kb = Kw / Ka = (1.0 × 10-14) / (4.9 × 10-10) = 2.04 × 10-5
This tells you cyanide is a weak base, but not an extremely weak one. A Kb on the order of 10-5 means measurable hydroxide will be produced.
Step 3: Set up the equilibrium expression
For the hydrolysis reaction:
CN– + H2O ⇌ HCN + OH–
Use an ICE table:
- Initial: [CN–] = 0.21, [HCN] = 0, [OH–] = 0
- Change: -x, +x, +x
- Equilibrium: [CN–] = 0.21 – x, [HCN] = x, [OH–] = x
The equilibrium expression is:
Kb = [HCN][OH–] / [CN–] = x2 / (0.21 – x)
Substitute Kb = 2.04 × 10-5:
2.04 × 10-5 = x2 / (0.21 – x)
Step 4: Solve for x, which equals [OH–]
There are two standard paths. The first is the weak-base approximation, which assumes x is small compared with 0.21. The second is the exact quadratic solution. For this problem, both methods give very similar answers.
- Approximation method: replace 0.21 – x with 0.21, so x ≈ √(Kb × C)
- Exact method: solve x2 + Kb x – KbC = 0 using the quadratic formula
Using the approximation:
x ≈ √[(2.04 × 10-5)(0.21)] ≈ 2.07 × 10-3 M
Using the exact quadratic gives a nearly identical value, about:
[OH–] = 2.06 × 10-3 M
That small difference is why many classroom settings accept the approximation. The percent ionization is less than 5 percent, so the small-x assumption is valid.
Step 5: Calculate pOH and pH
Now convert hydroxide concentration into pOH:
pOH = -log[OH–] = -log(2.06 × 10-3) ≈ 2.69
Then use:
pH = 14.00 – pOH = 14.00 – 2.69 = 11.31
Final answer: the pH of a 0.21 M NaCN solution is approximately 11.31 at 25 C.
Core constants and calculated values
| Quantity | Symbol | Value at 25 C | Why it matters |
|---|---|---|---|
| NaCN concentration | C | 0.21 M | Initial concentration of CN– |
| Acid dissociation constant of HCN | Ka | 4.9 × 10-10 | Used to find Kb of CN– |
| Ion product of water | Kw | 1.0 × 10-14 | Links Ka and Kb |
| Base dissociation constant of CN– | Kb | 2.04 × 10-5 | Controls hydroxide formation |
| Hydroxide concentration | [OH–] | 2.06 × 10-3 M | Direct output of hydrolysis equilibrium |
| pOH | pOH | 2.69 | Intermediate logarithmic step |
| pH | pH | 11.31 | Final answer |
How concentration changes the pH of NaCN solutions
One of the most useful ways to understand weak-base problems is to compare several concentrations while keeping Ka and temperature fixed. Because [OH–] for a weak base tends to scale roughly with the square root of concentration, pH rises as concentration rises, but not in a perfectly linear way.
| NaCN concentration (M) | Exact [OH–] (M) | pOH | pH at 25 C |
|---|---|---|---|
| 0.010 | 4.42 × 10-4 | 3.35 | 10.65 |
| 0.050 | 1.00 × 10-3 | 3.00 | 11.00 |
| 0.100 | 1.42 × 10-3 | 2.85 | 11.15 |
| 0.210 | 2.06 × 10-3 | 2.69 | 11.31 |
| 0.500 | 3.18 × 10-3 | 2.50 | 11.50 |
This data shows a realistic chemical trend: increasing NaCN concentration pushes the hydrolysis equilibrium toward more OH–, which raises pH. Still, the pH increase is modest because weak-base equilibria are governed by logarithms and by the square-root dependence from the equilibrium expression.
Common mistakes students make
- Assuming NaCN is neutral because it is a salt. Not all salts are neutral. Salts of strong bases and weak acids are basic.
- Using Ka directly as if NaCN were acidic. The reacting species is CN–, so you need Kb.
- Forgetting that Na+ is a spectator ion and does not control pH here.
- Mixing up pOH and pH. For a basic solution, you often find pOH first, then convert to pH.
- Using 14.00 for pH + pOH when the problem specifies a temperature other than 25 C. Strictly speaking, that sum changes with temperature because Kw changes.
When the small-x approximation is acceptable
The approximation x << C is commonly tested in chemistry courses. For weak bases, it generally works well when the degree of hydrolysis is small. In practice, after you estimate x, compare x/C. If it is less than about 5 percent, the approximation is usually acceptable. Here:
(2.06 × 10-3) / 0.21 ≈ 0.0098 = 0.98 percent
That is safely below 5 percent, so the approximation is valid. However, modern calculators can solve the quadratic instantly, so the exact approach is preferred for precision and for web calculators like the one above.
Why this problem matters in practical chemistry
Understanding the pH of cyanide-containing solutions matters beyond the classroom. Cyanide speciation depends strongly on pH because CN– and HCN interconvert. At lower pH, more hydrogen cyanide can form, and HCN is volatile and highly hazardous. In basic solution, the equilibrium shifts more toward CN–. That is one reason pH control is important in industrial, environmental, and analytical contexts involving cyanide compounds.
For credible background reading, consult authoritative sources such as the U.S. Environmental Protection Agency cyanide resources, the Centers for Disease Control and Prevention information on cyanide, and the NIST Chemistry WebBook. These sources help place the equilibrium calculation into a broader scientific and safety context.
Fast summary formula for this exact problem
- Start with 0.21 M NaCN, so initial [CN–] = 0.21 M.
- Use Ka(HCN) = 4.9 × 10-10.
- Compute Kb = Kw / Ka = 1.0 × 10-14 / 4.9 × 10-10 = 2.04 × 10-5.
- Write Kb = x2 / (0.21 – x).
- Solve for x = [OH–] ≈ 2.06 × 10-3 M.
- Find pOH = 2.69.
- Find pH = 11.31.
Final conclusion
To calculate the pH of a 0.21 M NaCN solution, treat cyanide as a weak base, not as a neutral spectator ion. The governing equilibrium is the hydrolysis of CN– with water. Using Ka for HCN and converting it to Kb for CN–, you find a hydroxide concentration of about 2.06 × 10-3 M. That gives a pOH of about 2.69 and a final pH of about 11.31 at 25 C.
If you want to explore how the answer changes with concentration, Ka, or temperature, the calculator above lets you adjust all of those inputs instantly. That makes it useful for homework checking, lab prework, and quick conceptual comparisons between weak-base salt solutions.