Calculate OH and pH for 0.10 M NaCN
Use this interactive sodium cyanide hydrolysis calculator to estimate hydroxide concentration, pOH, pH, and equilibrium concentrations for a 0.10 M NaCN solution. The default setup uses the acid dissociation constant of HCN at 25 degrees Celsius and applies the weak base equilibrium for CN– in water.
Default example: 0.10 M NaCN at 25 degrees Celsius, assuming complete dissociation into Na+ and CN–.
Calculator Results
How to calculate OH and pH for 0.10 M NaCN
To calculate OH and pH for 0.10 M NaCN, you treat sodium cyanide as a strong electrolyte that dissociates essentially completely into Na+ and CN– in water. The sodium ion is a spectator ion for acid base purposes, but cyanide is the conjugate base of hydrocyanic acid, HCN, which is a weak acid. Because CN– is a weak base, it reacts with water to generate hydroxide ions:
That reaction is the reason a sodium cyanide solution is basic. The key idea is that the hydroxide concentration does not come directly from the NaCN formula itself. Instead, it is generated by hydrolysis of the cyanide ion. Once you calculate the equilibrium hydroxide concentration, you can obtain pOH from the negative logarithm of [OH–] and then compute pH from the relationship pH = 14.00 – pOH at 25 degrees Celsius.
Step 1: Write the relevant equilibrium
For a 0.10 M sodium cyanide solution, the starting concentration of CN– is approximately 0.10 M. The equilibrium reaction is:
The base dissociation constant for cyanide is related to the acid dissociation constant of HCN:
If you use a common textbook value of Ka for HCN equal to 6.2 × 10-10 at 25 degrees Celsius, then:
Step 2: Set up an ICE table
An ICE table is often the clearest way to organize the calculation.
| Species | Initial (M) | Change (M) | Equilibrium (M) |
|---|---|---|---|
| CN– | 0.10 | -x | 0.10 – x |
| HCN | 0 | +x | x |
| OH– | 0 | +x | x |
Now substitute these equilibrium concentrations into the expression for Kb:
Step 3: Solve for x, which equals [OH-]
If you use the approximation method for a weak base, you assume x is small compared with 0.10. Then:
So the hydroxide concentration is approximately 1.27 × 10-3 M.
You can check whether the approximation is valid by comparing x with the initial concentration:
Since the percent ionization is below 5%, the approximation is acceptable. The exact quadratic method gives a very similar answer and is what this calculator can compute automatically if you choose that option.
Step 4: Convert [OH-] to pOH and pH
Once [OH–] is known, the remaining steps are straightforward:
Therefore, the pH of 0.10 M NaCN is approximately 11.10 at 25 degrees Celsius when Ka(HCN) = 6.2 × 10-10.
Quick answer for 0.10 M NaCN
- Initial CN– concentration: 0.10 M
- Ka of HCN at 25 degrees Celsius: 6.2 × 10-10
- Kb of CN–: 1.61 × 10-5
- [OH–] at equilibrium: about 1.26 × 10-3 to 1.27 × 10-3 M
- pOH: about 2.90
- pH: about 11.10
Why NaCN is basic in water
Sodium cyanide is often confusing to students because the formula itself does not contain OH–. However, basicity in aqueous chemistry does not require a hydroxide unit in the formula. What matters is whether the dissolved species can accept a proton from water. Cyanide can, because it is the conjugate base of a weak acid. The weaker the parent acid, the stronger its conjugate base tends to be. Since hydrocyanic acid is weak, CN– has a measurable tendency to pull a proton from water and generate OH–.
This is a classic example of salt hydrolysis. Salts derived from a strong base and a weak acid often give basic solutions. NaCN comes from NaOH, a strong base, and HCN, a weak acid. The sodium ion does not influence pH significantly, but cyanide does. In contrast, a salt like NaCl is neutral because Cl– is the conjugate base of a strong acid and is too weak to react meaningfully with water.
Exact solution versus approximation
For many classroom problems, the square root approximation is used because it is fast and usually accurate when Kb is small relative to the starting concentration. Still, the exact method is more rigorous. If you solve the quadratic exactly for:
with C = 0.10 and Kb = 1.61 × 10-5, you obtain x very close to 1.26 × 10-3 M. The difference between the approximate and exact values is tiny for this concentration. That means the approximation is not only convenient but also justified.
| Method | Formula Used | [OH–] for 0.10 M NaCN | pH | Comment |
|---|---|---|---|---|
| Approximation | x ≈ √(Kb × C) | 1.27 × 10-3 M | 11.10 | Very good when x is small relative to C |
| Exact quadratic | x = [-Kb + √(Kb2 + 4KbC)] / 2 | 1.26 × 10-3 M | 11.10 | Best for precision and validation |
Useful equilibrium data for cyanide calculations
When people search for how to calculate OH and pH for 0.10 M NaCN, they often need more than the final answer. They need a reliable reference point for equilibrium constants and water ionization. The table below collects widely used values relevant to standard introductory chemistry calculations.
| Quantity | Typical Value | Meaning for the calculation |
|---|---|---|
| Ka of HCN at 25 degrees Celsius | 6.2 × 10-10 | Determines how weak HCN is and therefore how basic CN– will be |
| pKa of HCN | About 9.21 | Equivalent way to express acidity strength |
| Kw at 25 degrees Celsius | 1.0 × 10-14 | Used to convert Ka to Kb with Kb = Kw / Ka |
| Kb of CN– | 1.61 × 10-5 | Directly controls OH– formation in NaCN solution |
| Calculated pH of 0.10 M NaCN | About 11.10 | Expected answer under standard conditions |
Common mistakes when calculating pH of sodium cyanide
- Treating NaCN as a strong base. NaCN is not like NaOH. It produces OH– through equilibrium hydrolysis, not complete release of hydroxide.
- Using Ka directly instead of converting to Kb. Since CN– acts as a base, you need Kb. If only Ka for HCN is given, convert it first using Kb = Kw / Ka.
- Forgetting that x equals [OH–]. In the ICE table, the amount of OH– formed is the same amount of HCN produced.
- Confusing pOH and pH. The hydroxide concentration gives pOH first. Then calculate pH from pH + pOH = 14.00 at 25 degrees Celsius.
- Ignoring temperature assumptions. The standard 14.00 relation is specific to 25 degrees Celsius unless a different Kw is used.
What happens if the concentration changes?
The pH of sodium cyanide does not increase linearly with concentration because pH is logarithmic and the equilibrium expression depends on the square root relationship for weak bases. If you decrease NaCN concentration by a factor of 10, the hydroxide concentration does not decrease by a full factor of 10. Instead, under the approximation, [OH–] scales with the square root of concentration. This is why weak base pH problems often show moderate pH shifts across large concentration changes.
For students, this means that memorizing one pH is less useful than understanding the method. Once you know how to derive Kb, set up the ICE table, and solve for x, you can handle any sodium cyanide concentration, not just 0.10 M.
Safety and chemical context
Cyanide chemistry is academically important, but sodium cyanide is also a highly hazardous substance in real life. That means conceptual understanding should be paired with chemical safety awareness. Cyanide compounds can interfere with cellular respiration and require strict handling controls in laboratory and industrial settings. If you are using this page for educational work, keep in mind that pH calculation is only one part of understanding the behavior of cyanide in aqueous systems. Speciation between CN– and HCN also matters because HCN can be volatile, especially in more acidic conditions.
For safety and toxicological background, consult authoritative references such as the CDC Agency for Toxic Substances and Disease Registry cyanide fact sheet, the U.S. EPA cyanide compounds technical overview, and university level chemistry material from Washington University in St. Louis Chemistry. These sources help place the equilibrium calculation in a broader scientific and safety context.
Worked example summary
Here is the entire calculation in compact form:
Final takeaway
If you need to calculate OH and pH for 0.10 M NaCN, the most reliable path is to recognize that cyanide is a weak base, convert Ka of HCN to Kb of CN–, solve the hydrolysis equilibrium, and then convert [OH–] to pOH and pH. Under standard conditions, the answer is a hydroxide concentration around 1.26 × 10-3 to 1.27 × 10-3 M and a pH of approximately 11.10. This calculator automates those steps while still showing the equilibrium logic behind the result.