Calculate the pH of a 2.49 m Solution of NaCN
This premium calculator determines the pH of sodium cyanide solution by treating cyanide as a weak base, converting molality to molarity when needed, and solving the hydrolysis equilibrium with either the exact quadratic method or the common weak-base approximation.
Enter the numerical concentration of sodium cyanide.
If molality is selected, the calculator converts to molarity using the density assumption below.
Used only when the concentration is entered as molality. If density is unknown, 1.00 g/mL gives a practical classroom estimate.
Default value is based on a standard room-temperature acid dissociation constant for hydrocyanic acid.
The exact method is recommended for accuracy.
Default molar mass of sodium cyanide.
pH versus NaCN concentration
The chart compares the target solution with a wider concentration range using the same constants and density assumptions.
Expert Guide: How to Calculate the pH of a 2.49 m Solution of NaCN
To calculate the pH of a 2.49 m solution of sodium cyanide, you need to recognize what NaCN does in water. Sodium cyanide is a soluble ionic compound, so it dissociates essentially completely into Na+ and CN–. The sodium ion is a spectator ion for acid-base chemistry, but the cyanide ion is the conjugate base of hydrocyanic acid, HCN. Because HCN is a weak acid, CN– is a weak base. That means a NaCN solution is basic, not neutral.
The key equilibrium is the hydrolysis of cyanide in water:
CN– + H2O ⇌ HCN + OH–
This reaction produces hydroxide ions, which increase pH. In many textbook problems, the notation 2.49 m means 2.49 molal, or 2.49 moles of solute per kilogram of solvent. However, acid-base equilibrium calculations are usually expressed in molarity because equilibrium constants in introductory chemistry are often used with molar concentrations. If density is not provided, instructors commonly make a practical approximation and treat a moderately concentrated aqueous solution as if molality and molarity are close enough for a classroom estimate. That is exactly why this calculator lets you choose whether to treat the entered value as molality or molarity, and if you use molality, it converts to an estimated molarity from density.
Step 1: Identify the relevant equilibrium constant
You usually start with the acid dissociation constant of hydrocyanic acid. A typical room-temperature value is:
| Quantity | Typical value at 25 C | Why it matters |
|---|---|---|
| pKa of HCN | 9.21 | Lets you convert directly to Ka |
| Ka of HCN | 6.17 × 10-10 | Acid strength of HCN |
| Kw | 1.00 × 10-14 | Needed to compute Kb |
| Kb of CN– | 1.62 × 10-5 | Direct base constant for the hydrolysis reaction |
| Molar mass of NaCN | 49.01 g/mol | Used when converting molality to molarity |
The relationship between conjugate acid and conjugate base is:
Kb = Kw / Ka
If pKa = 9.21, then Ka = 10-9.21 ≈ 6.17 × 10-10. Therefore:
Kb = (1.00 × 10-14) / (6.17 × 10-10) ≈ 1.62 × 10-5
Step 2: Convert 2.49 m to molarity if needed
Strictly speaking, 2.49 m is molality, not molarity. If no density is given, many worked examples approximate the concentration as 2.49 M. That approach gives a very good instructional answer here. If you want to be more careful, use this conversion formula:
M = (1000 × d × m) / (1000 + m × MM)
where d is density in g/mL, m is molality, and MM is molar mass in g/mol.
Using the default estimate of 1.00 g/mL:
- m = 2.49 mol/kg
- MM of NaCN = 49.01 g/mol
- m × MM = 2.49 × 49.01 ≈ 122.04 g
- M = (1000 × 1.00 × 2.49) / (1000 + 122.04) ≈ 2.22 M
If, instead, your teacher expects the common shortcut of treating 2.49 m as 2.49 M, the final pH changes only slightly. That is why many classroom answers land in the same narrow range near pH 11.8.
Step 3: Set up the ICE table for cyanide hydrolysis
For the base reaction:
CN– + H2O ⇌ HCN + OH–
If the initial formal concentration of cyanide is C, then the equilibrium values are:
- [CN–] = C – x
- [HCN] = x
- [OH–] = x
The equilibrium expression is:
Kb = x2 / (C – x)
At this point you have two paths:
- Approximation method: assume x is small, so C – x ≈ C
- Exact method: solve the quadratic equation
Step 4: Solve for hydroxide concentration
If you use the textbook approximation and treat the solution as 2.49 M:
x ≈ √(Kb × C) = √[(1.62 × 10-5)(2.49)] ≈ 6.35 × 10-3 M
Then:
- pOH = -log(6.35 × 10-3) ≈ 2.20
- pH = 14.00 – 2.20 ≈ 11.80
If you use the exact quadratic method with C = 2.49 M:
x = [-Kb + √(Kb2 + 4KbC)] / 2
This gives x ≈ 6.34 × 10-3 M, so pH remains approximately 11.80. The exact and approximate methods agree very closely because the fraction ionized is small relative to the initial cyanide concentration.
If you convert 2.49 m to about 2.22 M using the default density estimate of 1.00 g/mL, the pH is still about 11.78. That tiny shift is a useful reminder that the solution is strongly basic regardless of whether the problem is handled with the classroom shortcut or a more careful concentration conversion.
Final answer for the common classroom interpretation
For the standard interpretation used in many general chemistry problems, the pH of a 2.49 m solution of NaCN is approximately:
pH ≈ 11.80
Why NaCN produces a basic solution
Students sometimes wonder why a salt can change pH. The answer is that not all salts are neutral. A salt formed from a strong base and a weak acid generally gives a basic solution. Sodium cyanide comes from NaOH, a strong base, and HCN, a weak acid. Since Na+ does not hydrolyze appreciably but CN– does, the net effect is generation of OH–. This is the same logic used for salts like sodium acetate, except cyanide is a somewhat stronger base than acetate because HCN is a weaker acid than acetic acid.
Comparison table: how concentration changes the pH of NaCN
The following values use pKa(HCN) = 9.21 at 25 C and the exact weak-base equilibrium treatment. These numbers show how pH rises as sodium cyanide concentration increases.
| NaCN concentration treated as M | [OH–] from exact solution | pOH | pH |
|---|---|---|---|
| 0.010 M | 3.95 × 10-4 M | 3.40 | 10.60 |
| 0.100 M | 1.27 × 10-3 M | 2.90 | 11.10 |
| 1.00 M | 4.01 × 10-3 M | 2.40 | 11.60 |
| 2.22 M | 5.99 × 10-3 M | 2.22 | 11.78 |
| 2.49 M | 6.34 × 10-3 M | 2.20 | 11.80 |
| 3.00 M | 6.96 × 10-3 M | 2.16 | 11.84 |
Common mistakes to avoid
- Treating NaCN as a neutral salt. It is not neutral because CN– is the conjugate base of a weak acid.
- Using Ka directly instead of Kb. For cyanide hydrolysis, you need the base constant of CN–.
- Confusing molality and molarity. The difference may be small in some classroom problems, but it is still conceptually important.
- Forgetting to convert pOH to pH. The equilibrium gives [OH–], so pOH is found first.
- Ignoring units and temperature assumptions. Constants vary somewhat with temperature, so 25 C is the standard basis unless stated otherwise.
When is the approximation valid?
The approximation x << C is valid when the hydroxide produced is much smaller than the starting cyanide concentration. For 2.49 M NaCN, x is about 0.00634 M. Compared with 2.49 M, that is only about 0.25 percent. Because that percentage is far below the common 5 percent rule of thumb, the square-root approximation is highly reliable. In practical terms, both the exact and approximate methods produce the same reported pH to two decimal places.
Real chemistry context and safety note
Sodium cyanide is a highly hazardous chemical. While this page focuses on equilibrium calculations, cyanide salts require strict laboratory controls because they can release hydrogen cyanide under acidic conditions. This is one reason pH control matters in cyanide-containing systems: lower pH can shift equilibrium toward HCN, a volatile and extremely toxic species. For theoretical chemistry problems, you mainly need the conjugate acid-base relationship. In real life, you also need rigorous safety procedures, ventilation, segregation from acids, and institutional handling protocols.
Authoritative references for deeper study
- PubChem: Sodium cyanide compound record
- U.S. Environmental Protection Agency: Cyanide resources
- NIST Chemistry WebBook: Hydrogen cyanide data
Bottom line
If your chemistry assignment asks for the pH of a 2.49 m solution of NaCN, the expected answer is usually around 11.8. The reasoning is straightforward: cyanide is the conjugate base of the weak acid HCN, so it hydrolyzes water to form hydroxide. Using pKa(HCN) = 9.21 gives Kb ≈ 1.62 × 10-5, and solving the base equilibrium for a concentration near 2.49 gives pH ≈ 11.80. A more careful molality-to-molarity conversion changes the result only slightly, keeping the solution firmly in the basic range.
Educational use only. This page is for chemistry calculation practice and does not replace laboratory safety training, chemical hygiene plans, or institutional protocols for cyanide handling.