Calculate The Ph Of 0.50 M Nacn Solution

This premium calculator solves the pH of a sodium cyanide solution by treating cyanide as the conjugate base of hydrocyanic acid. Enter concentration and acid constant values, then calculate pH, pOH, hydroxide concentration, and equilibrium details instantly.

NaCN pH Calculator

How to Calculate the pH of 0.50 M NaCN Solution

To calculate the pH of a 0.50 M NaCN solution, you need to recognize that sodium cyanide is a salt made from a strong base, sodium hydroxide, and a weak acid, hydrocyanic acid. That means the sodium ion does not affect pH significantly, but the cyanide ion does. Cyanide acts as a weak base in water and reacts with water to produce hydroxide ions. Because hydroxide concentration increases, the solution becomes basic and the pH rises above 7.

The key hydrolysis reaction is:

CN⁻ + H₂O ⇌ HCN + OH⁻

This equilibrium is controlled by the base dissociation constant for cyanide, Kb. Since cyanide is the conjugate base of hydrocyanic acid, you do not usually look up Kb directly. Instead, you calculate it from the acid dissociation constant of HCN using the relationship:

Kb = Kw / Ka

At 25°C, Kw = 1.0 × 10^-14. A common textbook value for Ka(HCN) is 4.9 × 10^-10. Using that value gives:

Kb = (1.0 × 10^-14) / (4.9 × 10^-10) = 2.04 × 10^-5

Quick answer: For a 0.50 M NaCN solution, the pH is approximately 11.50 when Ka(HCN) = 4.9 × 10^-10 and Kw = 1.0 × 10^-14.

Step-by-Step Method

1. Write the hydrolysis reaction for cyanide in water.
2. Determine Ka for HCN.
3. Convert Ka to Kb using Kb = Kw / Ka.
4. Set up an ICE table for the weak base equilibrium.
5. Solve for [OH⁻] either with the square root approximation or the full quadratic equation.
6. Find pOH = -log[OH⁻].
7. Convert to pH using pH = 14.00 – pOH at 25°C.

ICE Table for 0.50 M NaCN

Assume sodium cyanide fully dissociates in water, so the initial cyanide concentration is 0.50 M.

• Initial: [CN⁻] = 0.50, [HCN] = 0, [OH⁻] = 0
• Change: -x, +x, +x
• Equilibrium: [CN⁻] = 0.50 – x, [HCN] = x, [OH⁻] = x

Now substitute into the base equilibrium expression:

Kb = [HCN][OH⁻] / [CN⁻] = x² / (0.50 – x)

With Kb = 2.04 × 10^-5, you can first try the standard weak-base approximation:

x ≈ √(Kb × C) = √(2.04 × 10^-5 × 0.50)

x ≈ √(1.02 × 10^-5) ≈ 3.19 × 10^-3 M

Then:

• [OH⁻] ≈ 3.19 × 10^-3 M
• pOH = -log(3.19 × 10^-3) ≈ 2.50
• pH = 14.00 – 2.50 = 11.50

The approximation is excellent here because the change in concentration is much smaller than the initial 0.50 M concentration. The percent ionization is well under 5%, which confirms the shortcut is valid for a classroom calculation.

Why NaCN Solutions Are Basic

Students sometimes wonder why a salt can change pH at all. The answer depends on the acid and base that formed the salt:

• A salt from a strong acid and strong base is usually neutral.
• A salt from a weak acid and strong base is basic.
• A salt from a strong acid and weak base is acidic.
• A salt from a weak acid and weak base may be acidic, basic, or nearly neutral depending on the relative values of Ka and Kb.

NaCN belongs to the second category. The sodium ion, Na⁺, comes from the strong base NaOH and does not hydrolyze enough to matter. The cyanide ion, CN⁻, is the conjugate base of weak acid HCN, so it removes protons from water and generates hydroxide.

Comparison Table: pH of NaCN at Different Concentrations

The table below uses Ka(HCN) = 4.9 × 10^-10 and Kw = 1.0 × 10^-14. Values are based on the weak-base calculation and closely match exact equilibrium solutions at these concentrations.

NaCN Concentration (M)	Kb for CN⁻	Approximate [OH⁻] (M)	pOH	pH
0.010	2.04 × 10^-5	4.52 × 10^-4	3.34	10.66
0.050	2.04 × 10^-5	1.01 × 10^-3	3.00	11.00
0.10	2.04 × 10^-5	1.43 × 10^-3	2.84	11.16
0.50	2.04 × 10^-5	3.19 × 10^-3	2.50	11.50
1.00	2.04 × 10^-5	4.52 × 10^-3	2.34	11.66

Comparison Table: Common Salt Hydrolysis Outcomes

Salt	Parent Acid	Parent Base	Expected Solution Behavior	Reason
NaCl	HCl, strong	NaOH, strong	Approximately neutral	Neither ion hydrolyzes significantly
NH4Cl	HCl, strong	NH3, weak	Acidic	NH4+ donates protons to water
CH3COONa	CH3COOH, weak	NaOH, strong	Basic	Acetate hydrolyzes to make OH⁻
NaCN	HCN, weak	NaOH, strong	Basic	CN⁻ hydrolyzes strongly enough to raise pH

Exact Versus Approximate Solution

In many chemistry problems, the square root approximation is accepted:

x ≈ √(KbC)

However, advanced students and instructors may prefer the exact quadratic equation. Starting from:

Kb = x² / (C – x)

Rearrange to:

x² + Kb x – Kb C = 0

For C = 0.50 and Kb = 2.04 × 10^-5, the positive quadratic root gives nearly the same hydroxide concentration as the approximation. That is why this calculator uses the exact solution while also showing chemistry students the approximation logic in the explanatory content.

Important Assumptions in This Calculation

• The solution is dilute enough that activities are approximated by concentrations.
• The salt dissociates completely into Na⁺ and CN⁻.
• The calculation is performed at or near 25°C unless a custom Kw is entered.
• There are no other acids, bases, or buffers present in the mixture.

Common Mistakes to Avoid

1. Treating NaCN like a neutral salt. It is not neutral because cyanide is a basic anion.
2. Using Ka directly to calculate pH. You must first convert to Kb for cyanide.
3. Forgetting pOH. Hydroxide concentration gives pOH first, then pH.
4. Ignoring significant digits. Chemistry homework and lab reports often expect pH values reported to two decimal places when supported by the data.
5. Assuming all salts behave the same way. Salt hydrolysis depends completely on the acid-base strength of the parent species.

Worked Summary for 0.50 M NaCN

Here is the complete compact solution:

1. Ka(HCN) = 4.9 × 10^-10
2. Kb(CN⁻) = 1.0 × 10^-14 / 4.9 × 10^-10 = 2.04 × 10^-5
3. CN⁻ + H₂O ⇌ HCN + OH⁻
4. x ≈ √(2.04 × 10^-5 × 0.50) = 3.19 × 10^-3
5. pOH = -log(3.19 × 10^-3) = 2.50
6. pH = 14.00 – 2.50 = 11.50

So, the pH of a 0.50 M NaCN solution is about 11.50. If your textbook uses a slightly different value for Ka(HCN), your answer may vary by a few hundredths of a pH unit, but it will still be strongly basic.

Authority Sources for Further Study

• University-level chemistry resources on acid-base equilibria and salt hydrolysis
• U.S. Environmental Protection Agency information related to aqueous chemistry concepts
• NIST Chemistry WebBook for chemical reference data

Final Takeaway

If you need to calculate the pH of 0.50 M NaCN solution, the essential idea is simple: cyanide is the conjugate base of a weak acid, so it reacts with water to produce hydroxide ions. Use the hydrolysis equilibrium, convert Ka to Kb, solve for [OH⁻], then convert pOH to pH. For the common constant set used in general chemistry, the final pH is approximately 11.50.