Calculate the Solubility of Co(OH)2 at pH 11.50
Use this cobalt(II) hydroxide solubility calculator to estimate molar solubility, dissolved cobalt concentration, hydroxide concentration, and mass solubility at a fixed pH. This tool uses the Ksp relation for Co(OH)2(s) ⇌ Co2+ + 2OH- and is optimized for common-ion conditions at alkaline pH.
How to calculate the solubility of Co(OH)2 at pH 11.50
Cobalt(II) hydroxide, written as Co(OH)2, is a sparingly soluble ionic solid. When it contacts water, a small amount dissolves according to the equilibrium reaction Co(OH)2(s) ⇌ Co2+ + 2OH-. The equilibrium constant for this process is the solubility product constant, Ksp. If the solution pH is known, then the hydroxide ion concentration can be found directly, and from there the dissolved cobalt concentration can be estimated. At pH 11.50, the solution is strongly basic, so hydroxide is already present in substantial concentration. That common ion suppresses the dissolution of Co(OH)2 dramatically.
The standard chemistry workflow is straightforward. First, convert pH to pOH using pOH = 14.00 – pH, assuming dilute aqueous solution at 25°C. For pH 11.50, the pOH is 2.50. Next, compute hydroxide concentration: [OH-] = 10^-pOH = 10^-2.50 = 3.16 × 10^-3 M. Then use the Ksp expression for cobalt hydroxide:
If pH is fixed externally by a buffer or strong base, [OH-] is treated as known and essentially constant. In that case the molar solubility of Co(OH)2 is approximately equal to [Co2+], so:
Using Ksp = 5.92 × 10^-15 and [OH-] = 3.16 × 10^-3 M, the result is:
That means the dissolved cobalt ion concentration is extremely low at pH 11.50. If you want the answer in mass terms, multiply by the molar mass of Co(OH)2, about 92.947 g/mol:
So the solubility is about 5.92 × 10^-10 M, or approximately 5.50 × 10^-8 g/L of Co(OH)2 under the fixed-pH assumption. In microgram units, that is around 0.055 µg/L. This result illustrates why many metal hydroxides precipitate at elevated pH: hydroxide concentration strongly drives the equilibrium toward the solid.
Why pH 11.50 makes cobalt hydroxide much less soluble
The reason is the common-ion effect. In pure water, a sparingly soluble hydroxide dissolves until enough hydroxide accumulates to satisfy the Ksp condition. But if the surrounding solution already contains hydroxide because the pH is high, the equilibrium is pushed backward. Le Châtelier’s principle predicts that adding one of the products of dissolution, here OH-, reduces the amount of solid that can remain dissolved.
The square on hydroxide in the Ksp expression matters greatly. Since Ksp = [Co2+][OH-]², doubling hydroxide does not just halve the cobalt concentration; it reduces [Co2+] by a factor of four, all else equal. Increasing pH by 1.0 unit decreases pOH by 1.0 and increases [OH-] by a factor of 10. Because hydroxide is squared, the solubility of Co(OH)2 drops by about a factor of 100 for each one-unit rise in pH in the common-ion regime.
| pH | pOH | [OH-] (M) | Calculated Solubility, s (M) | Solubility (g/L) |
|---|---|---|---|---|
| 9.00 | 5.00 | 1.00 × 10^-5 | 5.92 × 10^-5 | 5.50 × 10^-3 |
| 10.00 | 4.00 | 1.00 × 10^-4 | 5.92 × 10^-7 | 5.50 × 10^-5 |
| 11.50 | 2.50 | 3.16 × 10^-3 | 5.92 × 10^-10 | 5.50 × 10^-8 |
| 12.50 | 1.50 | 3.16 × 10^-2 | 5.92 × 10^-12 | 5.50 × 10^-10 |
| 13.50 | 0.50 | 3.16 × 10^-1 | 5.92 × 10^-14 | 5.50 × 10^-12 |
This table highlights how sensitive cobalt hydroxide solubility is to pH. From pH 9.0 to 11.5, the solubility falls by five orders of magnitude. That is why pH adjustment is such a powerful tool in water treatment, analytical chemistry, and metal separation systems.
Step-by-step method used by the calculator
- Read the entered pH, Ksp, and molar mass.
- Compute pOH from 14.00 – pH.
- Convert pOH to hydroxide concentration using [OH-] = 10^-pOH.
- Apply the Ksp equation for Co(OH)2: Ksp = [Co2+][OH-]².
- Assume the dissolved cobalt concentration equals the molar solubility of Co(OH)2 under fixed pH conditions.
- Convert molar solubility to g/L, mg/L, and µg/L for practical interpretation.
- Generate a chart showing how solubility changes across the selected pH range.
For a fixed-pH system, this approach is the standard and most useful one. It matches laboratory and process situations where pH is controlled by a large external reservoir of acid or base. In such a case, the contribution of the dissolving solid to the total hydroxide balance is negligible compared with the buffer or bulk base already in solution.
Exact chemistry context and the main assumption
Whenever you calculate the solubility of a hydroxide at a stated pH, you are usually being asked for the equilibrium concentration of dissolved metal ion in a solution whose pH is already established. That means the pH acts as an imposed condition, not as something created solely by the dissolution of the solid. This distinction is important.
If Co(OH)2 were dissolving in pure water with no added base, the problem would be different. You would have to solve a coupled equilibrium because the dissolving solid itself creates hydroxide. Under those conditions, setting solubility to s gives [Co2+] = s and [OH-] = 2s, so Ksp = s(2s)² = 4s³. But at pH 11.50, the hydroxide concentration from the environment, 3.16 × 10^-3 M, is far larger than 2s, which here is only 1.18 × 10^-9 M. That huge difference justifies the common-ion approximation extremely well.
| Quantity | Value at pH 11.50 | Interpretation |
|---|---|---|
| External [OH-] | 3.16 × 10^-3 M | Hydroxide supplied by the basic solution |
| Calculated solubility, s | 5.92 × 10^-10 M | Equal to dissolved Co2+ under the fixed-pH model |
| Hydroxide produced by dissolution, 2s | 1.18 × 10^-9 M | Negligible compared with the external hydroxide level |
| Ratio of external OH- to generated OH- | About 2.67 × 10^6 | Confirms the approximation is highly valid |
Practical meaning of the answer
A molar solubility of 5.92 × 10^-10 M is exceptionally low. In environmental systems, plating operations, battery materials handling, and precipitation-based treatment processes, a pH this high strongly limits dissolved cobalt from a simple hydroxide equilibrium perspective. In plain language, once the pH reaches 11.50, cobalt(II) hydroxide is much more likely to remain as a solid precipitate than to stay dissolved.
However, real systems can be more complicated than the ideal Ksp model. Ligands such as ammonia, carbonate, citrate, EDTA, or natural organic matter may complex Co2+ and increase apparent solubility. Ionic strength and temperature can shift the effective equilibrium constant. Oxidation state changes may also matter, because cobalt chemistry can involve Co2+ and Co3+ under different conditions. The calculator therefore gives a rigorous baseline estimate, but not a full speciation model.
When this calculation is most reliable
- The pH is controlled or measured reliably.
- The solution is not dominated by strong complexing ligands.
- The temperature is near 25°C.
- You need a classical equilibrium estimate based on Ksp.
- The ionic strength is moderate enough that concentration approximates activity reasonably well.
Common mistakes when calculating Co(OH)2 solubility
- Using pH directly as [H+] or [OH-] without converting through pOH.
- Forgetting that hydroxide is squared in the Ksp expression.
- Using the wrong stoichiometry for Co(OH)2.
- Ignoring unit conversions when reporting g/L, mg/L, or µg/L.
- Applying the simple Ksp model to a system with strong chelators or unusual ionic strength without caution.
Worked answer for pH 11.50
Numerical summary
- pH = 11.50
- pOH = 2.50
- [OH-] = 3.16 × 10^-3 M
- Ksp = 5.92 × 10^-15
- Solubility of Co(OH)2 = 5.92 × 10^-10 M
- Mass solubility = 5.50 × 10^-8 g/L
- Mass solubility = 5.50 × 10^-5 mg/L
- Mass solubility = 0.0550 µg/L
If your assignment, report, or treatment design asks for “the solubility of Co(OH)2 at pH 11.50,” this is the standard answer format you should present. Include the equilibrium expression, show how you converted pH to hydroxide concentration, and state your assumption that pH is fixed externally.
Authoritative references for further reading
Note: Ksp values in literature can vary slightly depending on data source, temperature, and conventions. This calculator uses the value entered by the user, with a default of 5.92 × 10^-15 at 25°C for a standard educational estimate.