Calculate the pH of a 2.00 M NH4CN Solution
This ultra-premium calculator solves the acid-base equilibrium for ammonium cyanide, a salt formed from the weak acid HCN and the weak base NH3. Use the exact equilibrium solver or the fast classroom approximation to estimate the pH and species distribution at 25 degrees Celsius.
NH4CN pH Calculator
NH4+ + H2O ⇌ NH3 + H3O+CN- + H2O ⇌ HCN + OH-Approximation for equimolar weak-acid/weak-base salts: pH ≈ 0.5 × (pKa of NH4+ + pKa of HCN)
Expert guide: how to calculate the pH of a 2.00 M NH4CN solution
To calculate the pH of a 2.00 M NH4CN solution, you need to recognize that ammonium cyanide is not a simple neutral salt. It is made from NH4+, the conjugate acid of the weak base ammonia, and CN-, the conjugate base of the weak acid hydrogen cyanide. That means both ions react with water, and the final pH depends on which hydrolysis reaction is stronger. In this case, the cyanide ion is the slightly stronger hydrolyzing species, so the solution is basic, with a pH a little above 9.2 at 25 degrees Celsius using standard textbook constants.
The most efficient way to solve the problem is to compare the acid strength of NH4+ with the base strength of CN-. Once you know those relative strengths, you can predict whether the solution is acidic, basic, or very close to neutral. Then you can calculate the actual pH using either the common approximation or a more exact equilibrium treatment. This page gives you both, and the calculator above lets you test different constant values if your instructor or textbook uses a slightly different pKa for HCN or NH4+.
Step 1: Identify the ions and their acid-base roles
NH4CN dissociates essentially completely in water:
NH4CN(aq) → NH4+(aq) + CN-(aq)
Now look at each ion:
- NH4+ is a weak acid because it can donate a proton to water and form NH3.
- CN- is a weak base because it can accept a proton from water and form HCN.
Those hydrolysis reactions are:
- NH4+ + H2O ⇌ NH3 + H3O+
- CN- + H2O ⇌ HCN + OH-
Since one reaction produces hydronium and the other produces hydroxide, the pH depends on the competition between them.
Step 2: Convert known constants into the form you need
At 25 degrees Celsius, commonly used values are approximately:
- pKa of NH4+ = 9.25
- pKa of HCN = 9.21
- pKw = 14.00
From these values:
- Ka(NH4+) = 10-9.25 ≈ 5.62 × 10-10
- Ka(HCN) = 10-9.21 ≈ 6.17 × 10-10
- Kb(CN-) = Kw / Ka(HCN) ≈ 1.00 × 10-14 / 6.17 × 10-10 ≈ 1.62 × 10-5
Notice something important: although NH4+ is weakly acidic, CN- is the conjugate base of an extremely weak acid, so its hydrolysis is significant. That is why NH4CN solutions are basic overall.
| Species | Role in water | Typical constant at 25 degrees Celsius | Approximate value | Interpretation |
|---|---|---|---|---|
| NH4+ | Weak acid | Ka | 5.62 × 10-10 | Produces H3O+ by donating a proton |
| HCN | Weak acid | Ka | 6.17 × 10-10 | Very weak acid, so CN- is a relatively strong weak base |
| CN- | Weak base | Kb = Kw / Ka(HCN) | 1.62 × 10-5 | Produces OH- by taking a proton from water |
| H2O | Autoionization reference | Kw | 1.00 × 10-14 | Links acid and base constants at 25 degrees Celsius |
Step 3: Use the quick formula for a salt of a weak acid and a weak base
For a salt made from a weak acid and a weak base where both ions start at the same formal concentration, a very useful approximation is:
pH ≈ 7 + 0.5 log(Kb of base ion / Ka of acid ion)
For ammonium cyanide, that becomes:
pH ≈ 7 + 0.5 log(Kb(CN-) / Ka(NH4+))
Substitute the values:
pH ≈ 7 + 0.5 log((1.62 × 10-5) / (5.62 × 10-10))
The ratio is about 2.88 × 104. The common log of that ratio is about 4.459. Half of that is about 2.229. Therefore:
pH ≈ 7 + 2.229 = 9.229
So the pH of a 2.00 M NH4CN solution is approximately 9.23.
You can also express the same relationship in pKa form. Because the salt is equimolar in NH4+ and CN-, the approximation simplifies elegantly to:
pH ≈ 0.5 × (pKa of NH4+ + pKa of HCN)
Using 9.25 and 9.21:
pH ≈ 0.5 × (9.25 + 9.21) = 9.23
This is one of the cleanest ways to solve the problem on an exam.
Step 4: Why the 2.00 M concentration hardly changes the answer
Students often expect a concentrated solution to force the pH dramatically upward or downward. For strong acids or strong bases, concentration has a large effect. But for salts of a weak acid and weak base at equal stoichiometric concentration, the pH is often governed mainly by the ratio of hydrolysis strengths, not by the actual concentration. That is why the 2.00 M value matters much less than you might think.
In a more exact equilibrium treatment, the concentration is still present in the mass-balance equations, but because both hydrolyzing ions begin at the same formal concentration, their contributions largely offset one another. The result is that the exact pH remains very close to the shortcut answer. For NH4CN with standard constants, the exact calculation is still essentially 9.23.
| Scenario | Formal salt concentration | Assumed constants | Approximate pH | Chemical meaning |
|---|---|---|---|---|
| Dilute NH4CN | 0.010 M | pKa(NH4+) = 9.25, pKa(HCN) = 9.21 | 9.23 | Hydrolysis strengths still dominate |
| Moderate NH4CN | 0.100 M | Same constants | 9.23 | Nearly identical classroom result |
| Target problem | 2.00 M | Same constants | 9.23 | Basic because CN- hydrolysis slightly exceeds NH4+ acidity |
| Very concentrated textbook idealization | 5.00 M | Same constants | 9.23 | Approximation remains nearly unchanged under ideal assumptions |
Exact equilibrium method for more advanced chemistry
If you want the more rigorous method, write the total concentration relationships for the ammonia and cyanide families:
- [NH4+] + [NH3] = C
- [CN-] + [HCN] = C
Then use the acid dissociation expressions:
- Ka(NH4+) = [H+][NH3] / [NH4+]
- Ka(HCN) = [H+][CN-] / [HCN]
Finally impose charge balance:
- [H+] + [NH4+] = [OH-] + [CN-]
When these are solved together, the exact pH lands extremely close to the shortcut result for typical textbook constants. This is why instructors often encourage the approximation first. It saves time without sacrificing meaningful accuracy.
Common mistakes students make
- Treating NH4CN like a neutral salt. It is not neutral. It contains two ions that both hydrolyze.
- Using only NH4+ and ignoring CN-. That would incorrectly predict an acidic solution.
- Using only CN- and ignoring NH4+. That would overestimate the basicity.
- Confusing pKa of HCN with pKb of CN-. Always convert carefully using Kw.
- Forgetting that the shortcut works because the salt is equimolar. The simple formula is not universal for all mixed weak acid-base systems.
How NH4CN compares with related salts
Comparisons are useful because they show why ammonium cyanide lands in the basic range rather than near 7 or below 7:
- NH4Cl is acidic because Cl- is essentially neutral but NH4+ is a weak acid.
- NaCN is strongly basic relative to NH4CN because Na+ is neutral while CN- is basic.
- NH4CN sits between those two extremes because it contains both a weak acid and a weak base.
The actual balance is decided by the fact that HCN is such a weak acid. Therefore, its conjugate base CN- is strong enough, within the weak-base category, to push the pH above neutral even in the presence of NH4+.
Practical note on notation: 2.00 M versus 2.00 m
In formal chemistry writing, M means molarity and m means molality. Many homework problems casually type lowercase m even when they mean molarity. This calculator is built around the usual general-chemistry treatment of a 2.00 M NH4CN solution, because pH problems are typically handled in molarity unless the problem explicitly states molality and asks you to account for density and activity corrections. In idealized textbook work, that distinction does not change the main conceptual result here: the solution is basic and its pH is about 9.23.
Best final answer for the target problem
Using standard 25 degree Celsius constants:
- pKa(NH4+) ≈ 9.25
- pKa(HCN) ≈ 9.21
Then:
pH ≈ 0.5(9.25 + 9.21) = 9.23
Therefore, the pH of a 2.00 M NH4CN solution is approximately 9.23.
Authoritative references for deeper study
If you want to verify the underlying chemistry or review weak acid-base equilibria from trusted sources, these references are useful: