Calculate The Ph Of 0.01 M Solution Of Nh4Cn

Use this premium calculator to estimate the pH of ammonium cyanide by applying weak acid and weak base equilibrium relationships at 25°C. The default values reflect standard textbook constants for NH3 and HCN.

Expert Guide: How to Calculate the pH of 0.01 M Solution of NH4CN

To calculate the pH of a 0.01 M solution of NH4CN, you need to recognize that ammonium cyanide is a salt formed from a weak base and a weak acid. Specifically, NH4CN dissociates in water into NH4+ and CN-. The NH4+ ion is the conjugate acid of NH3, while CN- is the conjugate base of HCN. Because both ions undergo hydrolysis, the pH depends on the relative strengths of the acid and base, not simply on concentration alone.

This is a classic equilibrium problem in general chemistry, analytical chemistry, and introductory acid-base theory. Many students initially assume that a salt solution must be neutral, but that is only true for salts of strong acids and strong bases, such as NaCl. In contrast, NH4CN contains ions that both react with water. The cyanide ion is a significantly stronger base than the ammonium ion is an acid, so the solution is basic.

For a salt made from a weak base and a weak acid, a highly useful relation is: pH = 7 + 0.5 log(Kb of weak base / Ka of weak acid)

Step 1: Identify the parent weak base and weak acid

Ammonium cyanide comes from:

• NH3, a weak base with Kb ≈ 1.8 × 10^-5
• HCN, a weak acid with Ka ≈ 6.2 × 10^-10

The ions in solution are:

• NH4+, which can donate H+ to water and act as a weak acid
• CN-, which can accept H+ from water and act as a weak base

Step 2: Apply the salt formula

At 25°C, a standard textbook shortcut for salts of weak acids and weak bases is:

pH = 7 + 0.5 log(Kb / Ka)

Substitute the values for NH3 and HCN:

pH = 7 + 0.5 log((1.8 × 10^-5) / (6.2 × 10^-10))

Now compute the ratio:

(1.8 × 10^-5) / (6.2 × 10^-10) ≈ 2.903 × 10^4

Take the logarithm:

log(2.903 × 10^4) ≈ 4.463

Then:

pH = 7 + 0.5(4.463) = 7 + 2.2315 = 9.23

Final answer: the pH of a 0.01 M solution of NH4CN is approximately 9.23.

Why the concentration often does not dominate this result

One detail that confuses many learners is that the concentration given in the problem is 0.01 M, but the simple formula above does not explicitly include concentration. This happens because for salts derived from a weak acid and a weak base, the dominant pH approximation depends mainly on the ratio of Kb to Ka. As long as the solution is not extremely dilute, the concentration terms tend to cancel in the derivation. That is why a 0.01 M NH4CN solution and a moderately similar concentration often produce nearly the same predicted pH by the shortcut method.

That said, concentration still matters in a full rigorous treatment, especially at very low concentrations where water autoionization becomes significant. In practical classroom chemistry, however, 0.01 M is concentrated enough that the standard approximation is generally appropriate.

Hydrolysis viewpoint for NH4CN

You can also understand this problem by examining each ion separately.

Ammonium ion as an acid

The ammonium ion reacts with water as follows:

NH4+ + H2O ⇌ NH3 + H3O+

Its acid dissociation constant is related to the base dissociation constant of NH3:

Ka(NH4+) = Kw / Kb(NH3)

Using Kw = 1.0 × 10^-14 and Kb(NH3) = 1.8 × 10^-5:

Ka(NH4+) ≈ 5.56 × 10^-10

Cyanide ion as a base

The cyanide ion hydrolyzes in water as:

CN- + H2O ⇌ HCN + OH-

Its base dissociation constant is related to the acid dissociation constant of HCN:

Kb(CN-) = Kw / Ka(HCN)

Using Ka(HCN) = 6.2 × 10^-10:

Kb(CN-) ≈ 1.61 × 10^-5

Notice the important comparison: Kb(CN-) is much larger than Ka(NH4+). This means the basic hydrolysis of cyanide dominates over the acidic hydrolysis of ammonium. Therefore, the solution ends up basic, with pH above 7.

Species	Role in Water	Equilibrium Constant	Approximate Value at 25°C
NH3	Weak base	Kb	1.8 × 10^-5
HCN	Weak acid	Ka	6.2 × 10^-10
NH4+	Weak acid in hydrolysis	Ka = Kw / Kb(NH3)	5.56 × 10^-10
CN-	Weak base in hydrolysis	Kb = Kw / Ka(HCN)	1.61 × 10^-5

Worked example for 0.01 M NH4CN

1. Write the parent weak base and weak acid: NH3 and HCN.
2. Look up or use standard constants: Kb(NH3) = 1.8 × 10^-5 and Ka(HCN) = 6.2 × 10^-10.
3. Apply the formula pH = 7 + 0.5 log(Kb/Ka).
4. Compute the ratio 1.8 × 10^-5 / 6.2 × 10^-10 = 2.903 × 10^4.
5. Take the common logarithm, giving about 4.463.
6. Multiply by 0.5 to obtain 2.2315.
7. Add this to 7 and get pH ≈ 9.23.

Comparison with other salt solutions

It helps to compare NH4CN with more familiar salts. Salts of strong acid plus strong base are neutral. Salts of strong base plus weak acid are basic. Salts of weak base plus strong acid are acidic. NH4CN belongs to the more subtle category of weak base plus weak acid, where the stronger hydrolysis reaction wins.

Salt	Parent Acid	Parent Base	Expected Solution Character	Typical pH Trend
NaCl	HCl strong acid	NaOH strong base	Neutral	About 7
NH4Cl	HCl strong acid	NH3 weak base	Acidic	Less than 7
NaCN	HCN weak acid	NaOH strong base	Basic	Greater than 7
NH4CN	HCN weak acid	NH3 weak base	Basic because CN- dominates	About 9.23

Common mistakes when solving NH4CN pH problems

• Treating NH4CN as neutral. It is not neutral because both ions hydrolyze.
• Using only NH4+ or only CN-. You must compare the acid strength of NH4+ with the base strength of CN-.
• Using Kb of CN- directly without deriving it. If only Ka of HCN is given, first compute Kb(CN-) from Kw/Ka.
• Ignoring the relationship between conjugate pairs. NH4+ and NH3 are conjugates, and HCN and CN- are conjugates.
• Forgetting that the formula uses Kb of the weak base and Ka of the weak acid. For this problem that means Kb of NH3 and Ka of HCN.

Deeper chemistry insight

The pH near 9.23 tells you that cyanide hydrolysis is strong enough to raise the hydroxide concentration significantly above pure water, but not so strong that the solution behaves like a strong base. This is exactly what you should expect from a weakly basic anion such as CN-. The ammonium ion partly offsets that basicity, but only slightly, because its conjugate acid strength is much smaller than the basic tendency of CN-.

Another subtle point is that equilibrium constants are temperature dependent. Most classroom pH calculations assume 25°C, where Kw is 1.0 × 10^-14. At other temperatures, both Kw and the listed Ka or Kb values can shift. If your instructor or laboratory manual provides a different temperature, you should use the matching constants.

Authority sources for equilibrium and acid-base constants

For reliable chemistry data and acid-base background, review these authoritative references:

• LibreTexts Chemistry for broad equilibrium derivations and worked examples.
• U.S. Environmental Protection Agency for cyanide chemistry context and environmental relevance.
• Princeton University Department of Chemistry for academic chemistry instruction resources.
• NIST Chemistry WebBook for vetted chemical property data.

Final takeaway

If your problem asks you to calculate the pH of a 0.01 M solution of NH4CN, the fastest correct route is to identify the parent weak base and weak acid, compare their strengths, and apply the weak acid plus weak base salt formula. With Kb(NH3) = 1.8 × 10^-5 and Ka(HCN) = 6.2 × 10^-10, the solution is basic, and the pH is about 9.23. This value makes chemical sense because CN- is a much stronger base than NH4+ is an acid.

Educational note: this calculator is designed for standard equilibrium practice and not for safety guidance or laboratory handling instructions. Cyanide compounds are hazardous and must only be handled under proper professional supervision and regulations.