Calculate OH and pH for 0.10 M NaCN
Use this interactive sodium cyanide hydrolysis calculator to determine hydroxide concentration, pOH, pH, percent hydrolysis, and equilibrium behavior for aqueous NaCN solutions.
NaCN pH Calculator
NaCN is the salt of a strong base and a weak acid. In water, CN– hydrolyzes to produce OH–, so the solution becomes basic.
Expert Guide: How to Calculate OH and pH for 0.10 M NaCN
If you need to calculate OH and pH for 0.10 M NaCN, the key idea is that sodium cyanide is not a neutral salt. It dissolves completely in water to produce Na+ and CN–, and the cyanide ion acts as a weak base. That basic behavior comes from the fact that CN– is the conjugate base of hydrocyanic acid, HCN, which is a weak acid. Since the conjugate acid is weak, the conjugate base is strong enough to react with water and generate hydroxide ions. That is why a 0.10 M NaCN solution has a pH well above 7.
The standard equilibrium for this problem is:
Sodium ions are spectator ions here, so they do not enter the acid-base equilibrium expression. The chemistry is controlled by cyanide hydrolysis. In most general chemistry settings, you are given or look up the acid dissociation constant of HCN, then convert it into a base dissociation constant for CN–. At 25 C, the relationship is:
Using a common textbook value of Ka(HCN) = 6.2 × 10^-10 and Kw = 1.0 × 10^-14, you get:
Because Kb is small but not tiny, a 0.10 M sodium cyanide solution produces measurable hydroxide. The exact amount can be estimated with the weak base shortcut or solved more rigorously with the quadratic equation. For this concentration, both approaches give nearly the same answer.
Step by Step Setup for 0.10 M NaCN
Start with the initial concentration of CN– as 0.10 M. Let x represent the amount that reacts with water:
Change: [CN-] = -x, [HCN] = +x, [OH-] = +x
Equilibrium: [CN-] = 0.10 – x, [HCN] = x, [OH-] = x
Substitute into the Kb expression:
For the approximation method, assume x is much smaller than 0.10:
x^2 = 1.61 × 10^-6
x = 1.27 × 10^-3 M
Since x is the hydroxide concentration produced by hydrolysis, we have:
pOH = -log(1.27 × 10^-3) = 2.90
pH = 14.00 – 2.90 = 11.10
Therefore, the commonly reported answer is pH ≈ 11.10 and [OH–] ≈ 1.27 × 10^-3 M for 0.10 M NaCN at 25 C using Ka = 6.2 × 10^-10 for HCN.
Why the Solution Is Basic
Students often ask why a salt such as NaCN changes pH while a salt like NaCl does not. The difference is the parent acid. Chloride is the conjugate base of hydrochloric acid, a strong acid, so Cl– has essentially no ability to pull a proton from water. Cyanide is the conjugate base of hydrocyanic acid, a weak acid, so it does react with water:
- NaCl: neutral in water because Cl– is negligibly basic.
- NaCN: basic in water because CN– hydrolyzes to form OH–.
- The stronger the conjugate base, the more the equilibrium shifts toward OH– production.
This is a classic strong base plus weak acid salt hydrolysis problem. The sodium ion does not contribute to pH, but cyanide absolutely does.
Exact Method Versus Approximation
In many chemistry courses, the square root shortcut is acceptable when the percent ionization or hydrolysis is small. For NaCN at 0.10 M, it is small enough to work. However, if you want a more rigorous answer, solve:
where C is the initial concentration. Using C = 0.10 M and Kb = 1.61 × 10^-5:
The exact result differs only slightly from the approximation, giving an [OH–] of about 1.26 × 10^-3 M and a pH still near 11.10. This tiny difference is why the approximation is commonly taught and accepted for this concentration.
| Method | Ka(HCN) | Kb(CN-) | [OH-] for 0.10 M NaCN | pOH | pH |
|---|---|---|---|---|---|
| Approximation | 6.2 × 10^-10 | 1.61 × 10^-5 | 1.27 × 10^-3 M | 2.90 | 11.10 |
| Exact quadratic | 6.2 × 10^-10 | 1.61 × 10^-5 | 1.26 × 10^-3 M | 2.90 | 11.10 |
| Approximation with alternate Ka | 4.9 × 10^-10 | 2.04 × 10^-5 | 1.43 × 10^-3 M | 2.85 | 11.15 |
How Sensitive Is the Answer to the Ka Value?
You may notice small differences in textbooks, databases, and lab manuals because HCN is often tabulated with slightly different Ka values. Some references report values near 4.9 × 10^-10, while others report values around 6.2 × 10^-10. Since pH depends on Kb, and Kb depends on Ka, your final pH can shift by a few hundredths. For educational work, this is completely normal. The important conclusion does not change: 0.10 M NaCN is clearly basic, and its pH is close to 11.1.
In practical terms:
- Choose the Ka value specified by your class or source.
- Convert Ka to Kb using Kw/Ka.
- Set up the hydrolysis equilibrium for CN–.
- Solve for x = [OH–].
- Find pOH, then convert to pH.
Comparison Table Across Several NaCN Concentrations
It helps to compare 0.10 M NaCN with weaker and stronger solutions. The values below use Ka(HCN) = 6.2 × 10^-10 at 25 C and the weak base approximation.
| NaCN concentration | Estimated [OH-] | pOH | pH | Percent hydrolysis |
|---|---|---|---|---|
| 0.0010 M | 1.27 × 10^-4 M | 3.90 | 10.10 | 12.7% |
| 0.010 M | 4.01 × 10^-4 M | 3.40 | 10.60 | 4.01% |
| 0.10 M | 1.27 × 10^-3 M | 2.90 | 11.10 | 1.27% |
| 1.0 M | 4.01 × 10^-3 M | 2.40 | 11.60 | 0.401% |
Common Mistakes When Solving NaCN pH Problems
- Using Ka directly instead of Kb. Because CN– is acting as a base, you need Kb for cyanide hydrolysis.
- Forgetting that Na+ is a spectator ion. Sodium does not control the pH in this problem.
- Confusing pOH with pH. Once you find [OH–], calculate pOH first, then use pH = 14 – pOH at 25 C.
- Assuming all salts are neutral. Salts of weak acids or weak bases often shift pH.
- Ignoring concentration effects. More concentrated NaCN solutions are generally more basic, although the percent hydrolysis decreases as concentration rises.
Safety and Real World Context
Cyanide chemistry is important not only in classroom acid-base equilibrium but also in industrial, mining, plating, and environmental contexts. Sodium cyanide is highly toxic, and solutions that contain cyanide must be handled using strict safety controls. In addition, pH management matters because acidic conditions can shift cyanide chemistry toward hydrogen cyanide gas, which is extremely hazardous. This is one reason pH monitoring is emphasized in professional cyanide handling guidance.
For authoritative background and chemistry resources, see:
- CDC NIOSH cyanide guidance
- PubChem sodium cyanide record from NIH
- Chemistry instructional resources used in university-level teaching
If you are working in an academic setting, compare the constants in your assigned text or instructor materials with reliable chemical databases or university references. Slight numerical differences in Ka are expected, but the hydrolysis method remains the same.
Quick Final Answer for 0.10 M NaCN
For the standard classroom problem, use Ka(HCN) = 6.2 × 10^-10 and Kw = 1.0 × 10^-14 at 25 C.
[OH-] ≈ 1.27 × 10^-3 M
pOH ≈ 2.90
pH ≈ 11.10
That means the solution is distinctly basic. If your source uses a slightly different Ka for HCN, your pH may shift to about 11.1 to 11.2, which is still consistent with the same underlying chemistry.
Best Practice for Exams and Lab Reports
When presenting your work, write the hydrolysis reaction, calculate Kb explicitly, show your ICE setup, and state whether you used the approximation or the exact quadratic. This makes your reasoning clear and protects you against losing points for unsupported answers. If you use the approximation, a quick validation step is useful:
Because the hydrolyzed fraction is comfortably below 5%, the approximation is justified. That is why the shortcut gives essentially the same pH as the exact method for this problem.