Calculating Ph Of 5M Solution Of Cn

Use this premium calculator to estimate the pH, pOH, hydroxide concentration, and percent ionization for a cyanide ion solution. By default, the tool uses a 5.00 M CN⁻ solution at 25°C and the relationship between cyanide and hydrocyanic acid, HCN.

Default chemistry CN⁻ + H₂O ⇌ HCN + OH⁻

Expected pH range at 5 M About 11.9 to 12.0

How to calculate the pH of a 5 M solution of CN⁻

Calculating the pH of a 5 M solution of CN⁻ is a classic weak-base equilibrium problem. The cyanide ion, written as CN⁻, is the conjugate base of hydrocyanic acid, HCN. Because HCN is a weak acid, its conjugate base is strong enough to react with water and produce hydroxide ions. That means a solution containing cyanide ion is basic, not neutral.

The key reaction is: CN⁻ + H₂O ⇌ HCN + OH⁻

Once you recognize that reaction, the task becomes straightforward. You convert the acid dissociation constant of HCN into the base dissociation constant of CN⁻, solve for the hydroxide concentration, determine pOH, and finally convert to pH. For a concentrated solution such as 5 M CN⁻, the result is strongly basic, and under standard 25°C assumptions it comes out near pH 11.95.

Important chemistry note: cyanide compounds can be extremely toxic. This page is for calculation and educational chemistry only, not handling or preparation guidance.

Step 1: Identify cyanide as a weak base

A common mistake is to look at a dissolved ionic species and assume it has no effect on pH. That is only true for ions that come from strong acids and strong bases and do not hydrolyze. Cyanide is different. Since HCN is a weak acid, CN⁻ can accept a proton from water:

• HCN is a weak acid with a small Ka.
• CN⁻ is its conjugate base.
• Conjugate bases of weak acids raise pH by generating OH⁻.

This is why a solution of sodium cyanide or potassium cyanide in water is basic. The identity of the spectator cation usually does not matter for the basicity calculation when the salt is fully dissociated and the cation is pH-neutral.

Step 2: Relate Ka of HCN to Kb of CN⁻

At 25°C, the relationship between conjugate acid-base constants is: Ka × Kb = Kw = 1.0 × 10⁻14

A commonly used value for hydrocyanic acid is: Ka(HCN) ≈ 6.2 × 10⁻10

Therefore: Kb(CN⁻) = 1.0 × 10⁻14 / 6.2 × 10⁻10 ≈ 1.61 × 10⁻5

If your textbook or instructor uses a slightly different Ka or pKa value for HCN, your final pH may differ by a few hundredths. That is normal in equilibrium chemistry because published constants are often rounded.

Step 3: Set up the ICE table

Assume an initial cyanide concentration of 5.00 M. Let x equal the concentration of OH⁻ produced at equilibrium:

• Initial: [CN⁻] = 5.00, [HCN] = 0, [OH⁻] = 0
• Change: [CN⁻] = -x, [HCN] = +x, [OH⁻] = +x
• Equilibrium: [CN⁻] = 5.00 – x, [HCN] = x, [OH⁻] = x

Then place those concentrations into the base equilibrium expression: Kb = [HCN][OH⁻] / [CN⁻] = x² / (5.00 – x)

Step 4: Solve for hydroxide concentration

Using Kb = 1.61 × 10⁻5:

1.61 × 10⁻5 = x² / (5.00 – x)

Because Kb is small relative to the formal concentration, many instructors first use the weak-base approximation: 5.00 – x ≈ 5.00

This gives: x ≈ √(KbC) = √((1.61 × 10⁻5)(5.00)) ≈ 8.97 × 10⁻3 M

So: [OH⁻] ≈ 8.97 × 10⁻3 M

If you solve the quadratic exactly instead of approximating, you obtain almost the same result: x ≈ 8.96 × 10⁻3 M

The approximation is valid here because x is tiny compared with 5.00 M. The percent change is only about 0.18%, well below the usual 5% rule used in general chemistry.

Step 5: Convert [OH⁻] to pOH and pH

Once hydroxide concentration is known: pOH = -log[OH⁻]

Using [OH⁻] ≈ 8.96 × 10⁻3 to 8.97 × 10⁻3 M:

• pOH ≈ 2.05
• pH = 14.00 – 2.05 ≈ 11.95

That is the standard answer for the pH of a 5 M cyanide ion solution under 25°C assumptions with Ka(HCN) near 6.2 × 10⁻10.

Worked example summary

1. Start with the equilibrium: CN⁻ + H₂O ⇌ HCN + OH⁻.
2. Use Kb = Kw / Ka.
3. If Ka(HCN) = 6.2 × 10⁻10, then Kb ≈ 1.61 × 10⁻5.
4. Set up Kb = x² / (5.00 – x).
5. Solve for x ≈ 8.96 × 10⁻3 M.
6. Compute pOH ≈ 2.05.
7. Compute pH ≈ 11.95.

Exact vs approximation data

In a weak-base problem, students often ask whether they must use the quadratic formula. For a 5 M CN⁻ solution, the approximation works very well. The table below compares approximate and exact calculations using Ka(HCN) = 6.2 × 10⁻10 and Kw = 1.0 × 10⁻14.

Method	Kb(CN⁻)	[OH⁻] at 5.00 M CN⁻	pOH	pH
Weak-base approximation	1.61 × 10⁻5	8.98 × 10⁻3 M	2.047	11.953
Exact quadratic solution	1.61 × 10⁻5	8.96 × 10⁻3 M	2.048	11.952
Difference	0	About 0.2%	About 0.001	About 0.001

The tiny difference explains why many textbooks accept either route as long as the setup is chemically correct. In high-precision work, however, the exact solution is still the best practice.

How concentration changes the pH of cyanide solutions

A 5 M cyanide solution is quite concentrated, so it is useful to compare it with more dilute solutions. Assuming the same equilibrium constants, pH rises as concentration rises because more cyanide is available to hydrolyze and generate OH⁻.

CN⁻ concentration (M)	Approximate [OH⁻] (M)	Approximate pOH	Approximate pH
0.001	1.27 × 10⁻4	3.896	10.104
0.010	4.01 × 10⁻4	3.397	10.603
0.100	1.27 × 10⁻3	2.896	11.104
1.00	4.01 × 10⁻3	2.397	11.603
5.00	8.98 × 10⁻3	2.047	11.953

These values are calculated using standard equilibrium relationships and show a sensible trend: pH increases with concentration, but not linearly. Because pH is logarithmic, each change in concentration produces a diminishing return in pH increase.

Why the answer is not as high as pH 14

Students sometimes expect a 5 M solution of a basic ion to have a pH near 14. That is not the case here because CN⁻ is not a strong base like hydroxide itself. It only partially reacts with water. Most of the cyanide remains as CN⁻ at equilibrium, and only a relatively small amount converts to HCN and OH⁻.

In other words, concentration alone does not guarantee an extreme pH. The equilibrium constant matters just as much. A concentrated weak base can still produce less hydroxide than a much less concentrated strong base.

Important assumptions behind the standard classroom result

When you calculate the pH of a 5 M solution of CN⁻ in general chemistry, you typically make several simplifying assumptions:

• Temperature is 25°C, so Kw = 1.0 × 10⁻14.
• The solution behaves ideally enough that concentrations approximate activities.
• The cyanide source fully dissociates into ions.
• Water autoionization is negligible compared with hydroxide produced by CN⁻ hydrolysis.
• No side reactions, complexation, oxidation, or gas loss are considered.

For introductory and many intermediate calculations, these assumptions are entirely appropriate. In advanced analytical chemistry, especially at very high ionic strength like 5 M, activity corrections may shift the effective pH somewhat. Still, the standard educational answer remains near 11.95.

Common mistakes when calculating pH of CN⁻

Using Ka directly instead of converting to Kb

Ka belongs to HCN, not to CN⁻ acting as a base. To solve the pH of cyanide ion in water, you need Kb. That conversion is one of the most important steps.

Forgetting to calculate pOH first

Since the reaction produces OH⁻, the most direct route is:

1. Find [OH⁻]
2. Calculate pOH
3. Use pH = 14 – pOH

Assuming cyanide is a strong base

CN⁻ is a weak base. Treating it as if all 5.00 M became OH⁻ would give a physically incorrect answer.

Dropping the quadratic without checking

The approximation is fine here, but that should be justified. Compare x with the initial concentration. If x is less than about 5% of the starting concentration, the approximation is generally considered acceptable.

Authoritative references for equilibrium and acid-base chemistry

If you want to verify acid-base relationships, pH concepts, and equilibrium theory from trusted educational or government sources, these references are helpful:

• LibreTexts Chemistry for broad acid-base and equilibrium explanations from educational institutions.
• U.S. Environmental Protection Agency for cyanide-related environmental chemistry context and safety information.
• NIST Chemistry WebBook for high-quality physical chemistry data resources from a U.S. government source.
• MIT Chemistry for educational chemistry materials from a leading university.

Final answer for the pH of a 5 M solution of CN⁻

Using standard aqueous equilibrium assumptions at 25°C and a typical value of Ka(HCN) ≈ 6.2 × 10⁻10, the cyanide ion has Kb ≈ 1.61 × 10⁻5. Solving the hydrolysis equilibrium gives an hydroxide concentration of about 8.96 × 10⁻3 M, which corresponds to:

• pOH ≈ 2.05
• pH ≈ 11.95

So the best concise answer is: the pH of a 5 M solution of CN⁻ is approximately 11.95.

Advanced note: at very high ionic strength, activity effects can matter. For educational problem solving, however, the accepted equilibrium-based classroom result remains approximately pH 11.95.