Calculate The Ph Of A 0.20 M Solution Of Kcn.

Use this premium weak-base salt hydrolysis calculator to determine the pH, pOH, hydroxide concentration, cyanide ion base constant, and a concentration-vs-pH chart for aqueous potassium cyanide at 25°C.

Salt of strong base + weak acid Default HCN Ka = 6.2 × 10^-10 Typical answer near pH 11.25

KCN dissociates into K⁺ and CN^–. Potassium is a spectator ion, while cyanide acts as a weak base by reacting with water to form HCN and OH^–.

Problem Focus

Weak base salt

Reference Temp

25°C

How to calculate the pH of a 0.20 M solution of KCN

To calculate the pH of a 0.20 M solution of potassium cyanide, you need to recognize the type of compound first. KCN is a salt formed from a strong base, KOH, and a weak acid, HCN. That means the potassium ion does not significantly affect pH, but the cyanide ion does. In water, CN^– behaves as a weak base and generates hydroxide ions, making the solution basic.

The key hydrolysis reaction is:

CN^– + H₂O ⇌ HCN + OH^–

Because OH^– is produced, the solution will have a pH above 7. The most important equilibrium constant is not the Ka of HCN directly, but the Kb of CN^–. However, Kb is easily found from the acid dissociation constant of HCN using the relationship:

K_b = K_w / K_a

At 25°C, K_w = 1.0 × 10^-14. If we use a typical value for HCN, K_a = 6.2 × 10^-10, then:

K_b = (1.0 × 10^-14) / (6.2 × 10^-10) = 1.61 × 10^-5

Set up the equilibrium expression

Start with the initial cyanide concentration of 0.20 M. Let x be the amount that reacts with water.

• Initial: [CN^–] = 0.20, [HCN] = 0, [OH^–] = 0
• Change: [CN^–] decreases by x, [HCN] increases by x, [OH^–] increases by x
• Equilibrium: [CN^–] = 0.20 – x, [HCN] = x, [OH^–] = x

The equilibrium expression becomes:

K_b = x² / (0.20 – x)

Substituting K_b = 1.61 × 10^-5 gives:

1.61 × 10^-5 = x² / (0.20 – x)

Approximation method

Since Kb is small relative to the initial concentration, most general chemistry courses allow the approximation 0.20 – x ≈ 0.20. Then:

x ≈ √(K_bC) = √[(1.61 × 10^-5)(0.20)] ≈ 1.80 × 10^-3 M

This x value equals the hydroxide concentration:

[OH^–] ≈ 1.80 × 10^-3 M

Now calculate pOH:

pOH = -log(1.80 × 10^-3) ≈ 2.75

And finally:

pH = 14.00 – 2.75 = 11.25

So the pH of a 0.20 M KCN solution is approximately 11.25 at 25°C.

Exact method

If you solve the quadratic equation instead of using the approximation, the answer changes only slightly. The exact result is essentially the same for most educational settings, confirming that the approximation is valid here. This is because the extent of hydrolysis is under 1% of the original 0.20 M concentration.

Final answer: For a 0.20 M aqueous KCN solution at 25°C, using Ka(HCN) = 6.2 × 10^-10, the calculated pH is about 11.25.

Why KCN makes the solution basic

Students often memorize that “strong acid plus strong base gives neutral” and “weak acid plus strong base gives basic,” but understanding why is much more useful. KCN contains K⁺, which comes from the strong base KOH, and CN^–, which is the conjugate base of the weak acid HCN. Strong-base cations generally do not hydrolyze enough to affect pH, but conjugate bases of weak acids often do.

The cyanide ion has enough basic character to abstract a proton from water. When it does, it forms HCN and OH^–. Since OH^– accumulates, the solution becomes alkaline. This is the same logic used for other salts such as sodium acetate, sodium fluoride, and sodium cyanide, though the exact pH depends on each conjugate base strength and concentration.

Conceptual checklist

1. Identify the ions produced when the salt dissolves.
2. Determine whether each ion comes from a strong or weak parent acid/base.
3. Ignore spectator ions like K⁺ from strong bases in introductory calculations.
4. Write the hydrolysis reaction for the ion that reacts with water.
5. Use K_w/K_a or K_w/K_b to convert between conjugate constants.
6. Solve for [OH^–] or [H⁺], then convert to pOH and pH.

Worked comparison table: how concentration changes pH for KCN

Because the hydroxide concentration for a weak base salt is roughly proportional to the square root of concentration, pH does not rise linearly with concentration. Doubling the KCN concentration does not double the pH change. The table below uses Ka(HCN) = 6.2 × 10^-10 and K_w = 1.0 × 10^-14 at 25°C.

KCN Concentration (M)	Kb of CN^–	Approximate [OH^–] (M)	Approximate pOH	Approximate pH
0.010	1.61 × 10^-5	4.02 × 10^-4	3.40	10.60
0.050	1.61 × 10^-5	8.98 × 10^-4	3.05	10.95
0.10	1.61 × 10^-5	1.27 × 10^-3	2.90	11.10
0.20	1.61 × 10^-5	1.80 × 10^-3	2.75	11.25
0.50	1.61 × 10^-5	2.84 × 10^-3	2.55	11.45

Exact vs approximate calculation for 0.20 M KCN

In many chemistry classes, you are expected to justify whether the small-x approximation is acceptable. A common rule is that if x is less than 5% of the initial concentration, the approximation is valid. For KCN at 0.20 M, x is about 0.0018 M, so:

% ionization = (0.0018 / 0.20) × 100 ≈ 0.90%

That is far below 5%, so the shortcut works very well. The exact and approximate values are therefore nearly identical.

Method	[OH^–] (M)	pOH	pH	Difference from Exact
Exact quadratic	1.787 × 10^-3	2.748	11.252	Reference
Approximation x ≈ √(KbC)	1.796 × 10^-3	2.746	11.254	About 0.002 pH units

Common mistakes when solving KCN pH problems

Even strong students can lose points on salt hydrolysis questions because of small setup errors. Here are the most common ones to watch for:

• Treating KCN like a neutral salt. It is not neutral because CN^– is the conjugate base of a weak acid.
• Using Ka directly instead of converting to Kb. The reacting species is CN^–, not HCN, so the base constant is needed.
• Forgetting that K⁺ is a spectator ion. Potassium does not appreciably hydrolyze in water.
• Using the strong-base formula. KCN is not a strong Arrhenius base like NaOH; it is a basic salt.
• Confusing pOH with pH. If you find [OH^–], calculate pOH first, then convert to pH.
• Dropping the temperature assumption. The common pH = 14 – pOH relationship is tied to K_w at 25°C in basic classroom problems.

How this calculation fits into general acid-base chemistry

The pH of KCN is a classic example of conjugate acid-base behavior. Once you understand this pattern, many related problems become easier. For example:

• NaF is basic because F^– is the conjugate base of weak acid HF.
• NH₄Cl is acidic because NH₄⁺ is the conjugate acid of weak base NH₃.
• NaCl is nearly neutral because both ions come from strong parent species.
• CH₃COONa is basic because acetate is the conjugate base of weak acetic acid.

So when you are asked to calculate the pH of a salt solution, the first question is not “what formula should I use?” but rather “which ion reacts with water?” That conceptual step usually determines the entire path to the answer.

Real-world and safety context

From a laboratory perspective, cyanide chemistry demands special caution. Potassium cyanide is highly toxic, and any educational discussion of KCN should remain strictly theoretical unless performed under approved professional safety controls. The pH calculation itself matters because cyanide speciation changes with acidity. In more acidic conditions, CN^– can shift toward molecular HCN, which is especially hazardous due to volatility and inhalation risk. That is one reason cyanide handling protocols emphasize pH control, ventilation, and formal institutional procedures.

In analytical and industrial contexts, acid-base equilibria influence cyanide behavior in extraction chemistry, electroplating processes, and environmental monitoring. Although a classroom pH problem is simplified, the same equilibrium ideas are directly relevant in professional settings. Understanding hydrolysis, conjugate pairs, and pH dependence is not just an academic exercise; it affects risk management and chemical process control.

Step-by-step summary for students

1. Write dissociation: KCN → K⁺ + CN^–
2. Ignore K⁺ as a spectator ion.
3. Write hydrolysis: CN^– + H₂O ⇌ HCN + OH^–
4. Calculate Kb: Kb = Kw / Ka = 1.0 × 10^-14 / 6.2 × 10^-10 = 1.61 × 10^-5
5. Use ICE setup with initial [CN^–] = 0.20 M
6. Solve for x, where x = [OH^–]
7. Find pOH = -log[OH^–]
8. Find pH = 14.00 – pOH
9. Report pH ≈ 11.25

Authoritative references and further reading

For academically reliable background on water chemistry, acid-base equilibria, and cyanide safety context, consult these sources:

• U.S. Environmental Protection Agency: Cyanide
• CDC NIOSH: Cyanide
• LibreTexts Chemistry: Acid Dissociation Constant and Related Equilibria

Bottom line

To calculate the pH of a 0.20 M solution of KCN, identify cyanide as a weak base, convert HCN’s Ka to the Kb of CN^–, solve the hydrolysis equilibrium, and convert the resulting hydroxide concentration to pH. Using standard 25°C values, the answer is approximately pH 11.25. The calculator above automates that process and also visualizes how pH changes with concentration near the selected value.