Calculate The Ph In A 0.00150M Ba Oh 2 Solution

Calculate the pH in a 0.00150 M Ba(OH)2 Solution

Use this interactive calculator to find hydroxide concentration, pOH, and final pH for barium hydroxide solutions. The default values are set for a 0.00150 M Ba(OH)2 solution at 25 C, where complete dissociation is assumed for this strong base.

Barium Hydroxide pH Calculator

Enter the analytical concentration of barium hydroxide in mol/L.
Ba(OH)2 dissociates to Ba2+ + 2OH.
This calculator uses the standard 25 C classroom relation.
Choose how many decimals to show in the results.
Default answer: For a 0.00150 M Ba(OH)2 solution, [OH] = 0.00300 M, pOH = 2.5229, and pH = 11.4771 at 25 C.

Visual Breakdown

The chart compares the input Ba(OH)2 concentration, the resulting hydroxide concentration, pOH, and pH.

How to calculate the pH in a 0.00150 M Ba(OH)2 solution

If you need to calculate the pH in a 0.00150 M Ba(OH)2 solution, the process is straightforward once you recognize that barium hydroxide is a strong base. In introductory and general chemistry, Ba(OH)2 is treated as dissociating completely in water under ordinary dilute conditions. That means each dissolved formula unit of barium hydroxide contributes one barium ion and two hydroxide ions. Because pH depends on hydrogen ion concentration and strong bases increase hydroxide ion concentration, the quickest path is to determine the total [OH] first, then compute pOH, and finally convert pOH to pH.

Final answer at 25 C: For 0.00150 M Ba(OH)2, the hydroxide concentration is 0.00300 M, the pOH is 2.5229, and the pH is 11.4771.

Step 1: Write the dissociation equation

Barium hydroxide is represented by the formula Ba(OH)2. When it dissolves in water, the idealized strong base dissociation is:

Ba(OH)2(aq) -> Ba2+(aq) + 2OH(aq)

This equation is the key to the entire calculation. The coefficient 2 in front of OH tells you that every mole of Ba(OH)2 yields two moles of hydroxide ions. Students often forget this stoichiometric multiplier and incorrectly use 0.00150 M as the hydroxide concentration. That would underestimate the basicity of the solution.

Step 2: Convert Ba(OH)2 molarity to hydroxide molarity

You are given:

  • Ba(OH)2 concentration = 0.00150 M
  • Hydroxide ions produced per formula unit = 2

So the hydroxide concentration is:

[OH] = 2 x 0.00150 = 0.00300 M

This is the most important numerical step. Once you have [OH], the rest follows directly from logarithms. Because 0.00300 M is much larger than 1.0 x 10-7 M, the contribution of hydroxide from pure water is negligible in this context.

Step 3: Calculate pOH

By definition:

pOH = -log[OH]

Substitute the hydroxide concentration:

pOH = -log(0.00300)

pOH = 2.5229

This value tells you how basic the solution is on the hydroxide scale. A lower pOH means a higher hydroxide concentration.

Step 4: Convert pOH to pH

At 25 C, the standard classroom relationship is:

pH + pOH = 14.00

Therefore:

pH = 14.00 – 2.5229 = 11.4771

Rounded appropriately, the pH is about 11.48. This confirms the solution is clearly basic.

Worked example summary

  1. Start with 0.00150 M Ba(OH)2.
  2. Recognize complete dissociation for a strong base.
  3. Multiply by 2 because each mole gives 2 moles of OH.
  4. Find [OH] = 0.00300 M.
  5. Compute pOH = -log(0.00300) = 2.5229.
  6. Use pH = 14.00 – 2.5229 = 11.4771.

Why Ba(OH)2 behaves this way

Barium hydroxide belongs to the group of ionic metal hydroxides that are considered strong bases in general chemistry. In solution, strong bases dissociate essentially completely, unlike weak bases such as ammonia, which require equilibrium calculations using Kb. That distinction matters. If this were a weak base problem, you would need an ICE table and a base dissociation constant. For Ba(OH)2, the stoichiometric ion count does almost all of the work.

It is also useful to remember that barium hydroxide produces more hydroxide per mole than sodium hydroxide. One mole of NaOH produces one mole of OH, while one mole of Ba(OH)2 produces two moles of OH. So, at the same analytical molarity, Ba(OH)2 creates a higher hydroxide concentration and a higher pH than NaOH.

Common mistakes students make

  • Forgetting the 2 in Ba(OH)2: The biggest error is treating [OH] as 0.00150 M instead of 0.00300 M.
  • Using pH = -log[OH]: That formula gives pOH, not pH.
  • Skipping units and significant figures: Molarity should be tracked, and logarithm answers should reflect the precision of the input data.
  • Confusing strong and weak bases: Ba(OH)2 is not solved with a weak base equilibrium setup in this kind of problem.
  • Using the 14.00 relation at nonstandard conditions without caution: In advanced chemistry, pH + pOH depends on temperature through Kw. For standard coursework, 25 C is usually assumed.

Comparison table: Ba(OH)2 versus other strong bases at the same formal concentration

The table below shows what happens when several common strong bases are present at 0.00150 M and are treated as fully dissociated at 25 C.

Base Formal concentration (M) OH produced per mole Resulting [OH] (M) pOH pH
NaOH 0.00150 1 0.00150 2.8239 11.1761
KOH 0.00150 1 0.00150 2.8239 11.1761
Ba(OH)2 0.00150 2 0.00300 2.5229 11.4771
Ca(OH)2 0.00150 2 0.00300 2.5229 11.4771

This comparison makes the stoichiometric point very clear. Bases with two hydroxide groups produce twice as much OH at the same molarity, which lowers pOH and raises pH. The numerical pH difference between a one hydroxide strong base and a two hydroxide strong base at 0.00150 M is about 0.3010 pH units, which comes directly from log(2).

Comparison table: pH as Ba(OH)2 concentration changes

Students often want to know how sensitive pH is to concentration. The table below uses the same strong base assumption and standard 25 C relationship.

Ba(OH)2 concentration (M) [OH] (M) pOH pH
0.000100 0.000200 3.6990 10.3010
0.00100 0.00200 2.6990 11.3010
0.00150 0.00300 2.5229 11.4771
0.0100 0.0200 1.6990 12.3010
0.100 0.200 0.6990 13.3010

These data show the logarithmic nature of the pH scale. A tenfold increase in Ba(OH)2 concentration changes pH by about 1.0 unit under these assumptions. Because pH is logarithmic, equal concentration changes do not produce equal pH changes unless they are multiplicative.

When the simple method is valid

For most homework, quiz, and exam problems involving a dilute but not extremely tiny concentration of Ba(OH)2, the simple method is exactly what your instructor expects. The assumptions are:

  • The base behaves as a strong electrolyte.
  • Dissociation is effectively complete.
  • The solution is dilute enough to ignore activity corrections.
  • Temperature is taken as 25 C so pH + pOH = 14.00.

At very high ionic strengths, a more advanced treatment could use activities rather than concentrations. At very low concentrations close to the autoionization of water, pure water contributions can become more relevant. Those are important in analytical chemistry and physical chemistry, but they are usually beyond the scope of a standard pH problem like this one.

Quick mental check for reasonableness

Whenever you compute pH, do a sanity check. A concentration of 0.00150 M is on the millimolar scale. Since Ba(OH)2 gives two hydroxides, the OH concentration is 0.00300 M. A hydroxide concentration around 10-3 should give a pOH near 3, and therefore a pH near 11. The exact answer, 11.4771, fits that expectation perfectly. If you got a pH near 3, 7, or 14.5, you would know immediately that something went wrong.

Authoritative chemistry references

For readers who want to verify core ideas such as water autoionization, pH scales, and general chemical data, these sources are useful:

For strict .gov or .edu examples relevant to chemistry fundamentals, you can also consult broad educational resources from agencies and universities. The calculation shown here follows the standard undergraduate approach used in introductory chemistry.

Final takeaway

To calculate the pH in a 0.00150 M Ba(OH)2 solution, multiply the base concentration by 2 to get the hydroxide concentration, use the negative logarithm to find pOH, and subtract from 14.00 to get pH. The complete calculation is:

[OH] = 2 x 0.00150 = 0.00300 M
pOH = -log(0.00300) = 2.5229
pH = 14.00 – 2.5229 = 11.4771

If you remember only one concept, remember this: Ba(OH)2 releases two hydroxide ions per formula unit. That single stoichiometric fact is what determines the correct pH.

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