Calculate the pH of a 0.010 m Solution of NaCN
Use this interactive calculator to estimate the pH, pOH, hydroxide concentration, and base hydrolysis behavior of sodium cyanide in water at 25 degrees Celsius.
NaCN pH Calculator
Results
Default estimate for 0.010 m NaCN using pKa(HCN) = 9.21 at 25 degrees Celsius.
How to calculate the pH of a 0.010 m solution of NaCN
To calculate the pH of a 0.010 m solution of sodium cyanide, you start by recognizing that NaCN is a salt made from a strong base, sodium hydroxide, and a weak acid, hydrocyanic acid. Because the sodium ion does not significantly affect pH in water, the chemistry is controlled by the cyanide ion, CN-. Cyanide behaves as a weak Brønsted base and reacts with water to produce hydroxide ions:
CN- + H2O ⇌ HCN + OH-
The formation of hydroxide means the solution is basic, so the pH will be greater than 7. The key constant is the base dissociation constant for cyanide, Kb. In most general chemistry settings, you are given or infer the acid dissociation constant of HCN first, because hydrocyanic acid is the conjugate acid of cyanide. The relationship between the two is:
Kb = Kw / Ka
At 25 degrees Celsius, the ionic product of water is Kw = 1.0 × 10^-14. A widely used pKa value for HCN is about 9.21, which corresponds to:
- Ka(HCN) = 10^-9.21 ≈ 6.17 × 10^-10
- Kb(CN-) = 1.0 × 10^-14 / 6.17 × 10^-10 ≈ 1.62 × 10^-5
Once you know Kb, you can set up an equilibrium table for a starting cyanide concentration of 0.010. Although the problem states 0.010 m, educational pH problems commonly treat very dilute aqueous molality and molarity as effectively the same for a quick estimate. That gives the equilibrium expression:
Kb = [HCN][OH-] / [CN-]
If x is the amount of CN- that reacts, then:
- [OH-] = x
- [HCN] = x
- [CN-] = 0.010 – x
Substituting into the equilibrium expression gives:
1.62 × 10^-5 = x^2 / (0.010 – x)
You can solve this either by approximation or with the quadratic formula. Because Kb is small but not extremely tiny relative to 0.010, the exact quadratic method is the cleaner and more defensible approach.
Step by step solution
- Write the base hydrolysis reaction: CN- + H2O ⇌ HCN + OH-
- Convert pKa of HCN to Ka: Ka = 10^-9.21 ≈ 6.17 × 10^-10
- Find Kb: Kb = 1.0 × 10^-14 / 6.17 × 10^-10 ≈ 1.62 × 10^-5
- Use the equilibrium relation: Kb = x^2 / (0.010 – x)
- Solve for x, where x = [OH-]
- Compute pOH = -log10[OH-]
- Compute pH = 14.00 – pOH
Using the exact quadratic solution:
x = (-Kb + √(Kb^2 + 4KbC)) / 2, where C = 0.010
This gives x ≈ 3.95 × 10^-4 M. Therefore:
- [OH-] ≈ 3.95 × 10^-4 M
- pOH ≈ 3.40
- pH ≈ 10.60
So, the pH of a 0.010 m solution of NaCN is approximately 10.60 when calculated with pKa(HCN) = 9.21 at 25 degrees Celsius.
Why NaCN makes water basic
Sodium cyanide is a classic example of a salt whose pH is controlled by the conjugate base of a weak acid. In water, NaCN dissociates essentially completely into sodium ions and cyanide ions. Sodium ions are spectators for acid-base purposes because they come from a strong base and have negligible tendency to react with water. Cyanide ions, however, are basic enough to remove protons from water molecules, producing hydroxide. That hydroxide is what pushes the pH upward.
The strength of this basicity depends on how weak HCN is as an acid. Since hydrocyanic acid is weak, its conjugate base is strong enough to hydrolyze water noticeably. That is why the pH is not only above 7 but substantially alkaline.
Key interpretation points
- If the conjugate acid is weaker, the conjugate base is stronger.
- If the salt concentration increases, the hydroxide concentration generally increases too.
- If temperature changes, Kw can change, and calculated pH may shift.
- If ionic strength becomes significant, activity corrections may matter in advanced work.
Approximation method versus exact quadratic method
Students are often taught the square root shortcut for weak acids and weak bases. For NaCN, that approximation is:
x ≈ √(Kb × C)
Using Kb = 1.62 × 10^-5 and C = 0.010:
- x ≈ √(1.62 × 10^-7) ≈ 4.03 × 10^-4
- pOH ≈ 3.39
- pH ≈ 10.61
That is very close to the exact answer because x is still only a few percent of the initial concentration. In many classroom settings, either answer is accepted if the assumptions are stated clearly. The calculator above allows you to compare both methods directly.
| Method | [OH-] from 0.010 NaCN | pOH | pH | Comment |
|---|---|---|---|---|
| Exact quadratic | 3.95 × 10^-4 M | 3.40 | 10.60 | Best general chemistry answer |
| Weak-base approximation | 4.03 × 10^-4 M | 3.39 | 10.61 | Fast and very close |
| Difference | About 2.0% | About 0.01 | About 0.01 | Negligible for most coursework |
Real data and reference values relevant to cyanide chemistry
Reliable calculations depend on sound reference constants. Different textbooks may round the pKa of HCN slightly differently, often giving values in the 9.2 range at room temperature. Those small differences can shift the final pH by a few hundredths. Below is a comparison table showing how that matters.
| Assumed pKa of HCN | Ka | Kb of CN- | Calculated pH for 0.010 concentration | Approximate shift |
|---|---|---|---|---|
| 9.20 | 6.31 × 10^-10 | 1.58 × 10^-5 | 10.60 | Reference point |
| 9.21 | 6.17 × 10^-10 | 1.62 × 10^-5 | 10.60 | Very small increase |
| 9.31 | 4.90 × 10^-10 | 2.04 × 10^-5 | 10.65 | About +0.05 pH units |
These values illustrate an important lesson: pH calculations are only as precise as the equilibrium constants and assumptions behind them. In introductory chemistry, reporting a pH of 10.6 is entirely appropriate. In analytical chemistry or environmental chemistry, you would state the reference constant source, temperature, and whether activities or concentrations were used.
Molality versus molarity in this problem
The notation 0.010 m technically means molality, or moles of solute per kilogram of solvent. Many pH equilibrium problems in introductory chemistry are solved using molarity because equilibrium expressions are traditionally written in terms of solution concentration. At very low concentration in water, 0.010 m and 0.010 M are close enough that the pH estimate changes only trivially, especially compared with normal uncertainty in equilibrium constants. That is why the calculator treats the value as effectively dilute aqueous concentration unless you are doing a more advanced thermodynamic activity analysis.
In formal thermodynamics, molality can be preferred because it is temperature independent, while molarity changes with volume. But for a typical homework or exam item phrased as “calculate the pH of a 0.010 m solution of NaCN,” the expected route is still the conjugate-base hydrolysis calculation shown above.
Common mistakes students make
- Using Ka directly instead of converting to Kb for CN-.
- Forgetting that Na+ is a spectator ion.
- Assuming the solution is neutral because NaCN is a salt.
- Calculating [H+] first even though the hydrolysis directly gives [OH-].
- Using pH = -log[OH-] instead of pOH = -log[OH-].
- Ignoring the need to subtract from 14 at 25 degrees Celsius.
- Rounding too early and losing meaningful precision.
When a more advanced treatment is needed
For concentrated solutions, nonideal solutions, or rigorous environmental calculations, chemists may use activities instead of simple concentrations. Ionic strength affects effective chemical behavior, and species such as HCN(aq), CN-, OH-, and dissolved CO2 can complicate the system. In air-exposed solutions, carbon dioxide can consume hydroxide and slightly alter the apparent pH. In real cyanide systems, safety, volatility of HCN, and metal complexation can also matter.
However, none of those complications changes the standard classroom answer for this problem. Under the usual assumptions, the pH is about 10.6.
Authoritative references for acid-base constants and aqueous chemistry
If you want to verify equilibrium constants or review aqueous acid-base theory from authoritative sources, these references are useful:
- U.S. Environmental Protection Agency: cyanide overview and aquatic chemistry context
- NIST Chemistry WebBook: thermochemical and chemical reference data
- Chemistry LibreTexts hosted by higher education institutions: acid-base equilibrium tutorials
Final answer
Assuming a dilute aqueous solution at 25 degrees Celsius and using pKa(HCN) = 9.21, the cyanide ion acts as a weak base with Kb ≈ 1.62 × 10^-5. Solving the hydrolysis equilibrium for a 0.010 concentration gives [OH-] ≈ 3.95 × 10^-4 M, pOH ≈ 3.40, and therefore:
pH ≈ 10.60