Calculate The Ph Of 0.0555M Boric Acid Ka 5.8X10-10

Use this interactive weak acid calculator to find the pH of boric acid from its molarity and acid dissociation constant. The default values are already set to solve the exact problem: calculate the pH of 0.0555 M boric acid with Ka = 5.8 × 10^-10.

How to calculate the pH of 0.0555 M boric acid when Ka = 5.8 × 10^-10

To calculate the pH of a weak acid solution, you use the acid dissociation equilibrium rather than assuming complete ionization. For the problem calculate the pH of 0.0555 M boric acid, Ka = 5.8 × 10^-10, the key idea is that boric acid is a very weak acid, so only a tiny fraction of the dissolved molecules generate hydrogen ions. That means the pH will be acidic, but not nearly as low as a strong acid of the same concentration.

In many introductory chemistry settings, boric acid is modeled as a monoprotic weak acid for pH calculations using the standard weak acid equilibrium approach. With that model, the dissociation can be represented in a simplified way as:

HA ⇌ H⁺ + A^–

Here, HA represents boric acid in the weak acid model, H⁺ is the hydrogen ion concentration generated by dissociation, and A^– is the conjugate base. The acid dissociation constant is:

Ka = [H⁺][A^–] / [HA]

Given:

• Initial acid concentration, C = 0.0555 M
• Ka = 5.8 × 10^-10

Let x be the amount of acid that dissociates. Then at equilibrium:

• [H⁺] = x
• [A^–] = x
• [HA] = 0.0555 – x

Substitute into the equilibrium expression:

5.8 × 10^-10 = x² / (0.0555 – x)

Because Ka is extremely small, x is much smaller than 0.0555, so the common weak acid approximation says:

5.8 × 10^-10 ≈ x² / 0.0555

Now solve for x:

x = √(Ka × C) = √((5.8 × 10^-10)(0.0555)) ≈ 5.67 × 10^-6 M

Since x is the hydrogen ion concentration:

pH = -log[H⁺] = -log(5.67 × 10^-6) ≈ 5.25

Final answer: The pH of 0.0555 M boric acid with Ka = 5.8 × 10^-10 is approximately 5.25.

Exact solution versus approximation

For high quality chemistry work, it is useful to check the exact quadratic solution. Starting from:

Ka = x² / (C – x)

Rearrange into a quadratic:

x² + Ka x – KaC = 0

Then solve:

x = (-Ka + √(Ka² + 4KaC)) / 2

Using C = 0.0555 and Ka = 5.8 × 10^-10, the exact value of x is still about 5.67 × 10^-6 M, so the pH remains about 5.25. The approximation works very well because the percent ionization is extremely small.

Why boric acid has a relatively high pH compared with stronger acids

A 0.0555 M solution may look fairly concentrated, but concentration alone does not determine pH. Acid strength matters just as much. Since the Ka value of boric acid is only 5.8 × 10^-10, it ionizes only slightly in water. This means most boric acid molecules remain undissociated at equilibrium, so the hydrogen ion concentration stays low.

That is the core reason the pH is around 5.25 instead of near 1 or 2. If this were a strong acid at 0.0555 M, the pH would be much lower because nearly all of the acid would release hydrogen ions.

Key takeaways

• Boric acid is a weak acid in this calculation framework.
• Its Ka is very small, so dissociation is limited.
• The hydrogen ion concentration is approximately 5.67 × 10^-6 M.
• The resulting pH is approximately 5.25.
• The exact quadratic and the square root approximation give essentially the same answer here.

Useful chemistry checks

• If x is much smaller than the initial concentration, the weak acid approximation is valid.
• If percent ionization is far below 5%, the approximation is usually safe.
• The pH of a weak acid solution rises as concentration decreases.
• The pH falls if Ka increases because the acid dissociates more strongly.

Step by step ICE table setup

An ICE table is one of the clearest ways to solve weak acid equilibrium problems. ICE stands for Initial, Change, and Equilibrium.

1. Initial: Start with 0.0555 M boric acid, and assume 0 M H⁺ and 0 M A^– from the acid itself.
2. Change: If x dissociates, boric acid decreases by x while H⁺ and A^– each increase by x.
3. Equilibrium: [HA] = 0.0555 – x, [H⁺] = x, [A^–] = x.
4. Substitute into Ka: 5.8 × 10^-10 = x² / (0.0555 – x)
5. Solve for x: x ≈ 5.67 × 10^-6 M
6. Convert to pH: pH = -log(x) ≈ 5.25

Comparison table: boric acid pH at several concentrations using Ka = 5.8 × 10^-10

The table below shows how pH changes as boric acid concentration changes, while keeping the same Ka value. These are calculated weak acid equilibrium values at standard conditions and help put the 0.0555 M case into context.

Initial concentration (M)	Estimated [H^+] (M)	Calculated pH	Interpretation
0.1000	7.62 × 10^-6	5.12	More concentrated solution, slightly lower pH
0.0555	5.67 × 10^-6	5.25	The requested problem result
0.0100	2.41 × 10^-6	5.62	Dilution reduces hydrogen ion concentration
0.0010	7.62 × 10^-7	6.12	Very dilute weak acid solution approaches neutral pH

Comparison table: percent ionization across concentrations

Another valuable statistic is percent ionization, which shows what fraction of the acid actually dissociates. For weak acids, this quantity is often very small.

Initial concentration (M)	[H^+] (M)	Percent ionization	What it tells you
0.1000	7.62 × 10^-6	0.0076%	Only a tiny fraction dissociates
0.0555	5.67 × 10^-6	0.0102%	Approximation is strongly justified
0.0100	2.41 × 10^-6	0.0241%	Ionization increases as concentration decreases
0.0010	7.62 × 10^-7	0.0762%	Still a weakly ionized acid, but relatively more dissociation

Common mistakes when solving this type of pH problem

• Treating boric acid like a strong acid: This would produce a much lower and incorrect pH.
• Forgetting the equilibrium expression: Weak acids must be solved with Ka, not full dissociation.
• Using pOH by mistake: Once you find [H⁺], take the negative logarithm directly to get pH.
• Ignoring scientific notation: Small errors in exponents can shift the pH significantly.
• Not checking the approximation: It is good practice to confirm that x is much smaller than the initial concentration.

How this calculator works

This calculator takes the input concentration and Ka value, then solves the weak acid equilibrium either with the exact quadratic formula or with the square root approximation. It returns the hydrogen ion concentration, pH, pOH, equilibrium weak acid concentration, conjugate base concentration, and percent ionization. It also plots the major concentrations in a bar chart so you can immediately see how little of the acid dissociates.

For the requested values, the chart makes the chemistry visually obvious: the equilibrium concentration of undissociated boric acid remains close to the original 0.0555 M, while the H⁺ and A^– concentrations are tiny in comparison.

Authoritative references for weak acid chemistry and water quality context

If you want to verify acid-base concepts or explore boron and water chemistry further, these authoritative resources are useful:

• U.S. Environmental Protection Agency: pH overview
• National Institutes of Health and NCBI: boron and boric acid background information
• Chemistry LibreTexts educational resource

Final conclusion

When asked to calculate the pH of 0.0555 M boric acid with Ka = 5.8 × 10^-10, the correct weak acid equilibrium approach gives a pH of about 5.25. The exact and approximate methods agree very closely because boric acid is extremely weak and dissociates only slightly at this concentration. That low degree of ionization is also why the percent ionization is only about 0.0102%.

If you are studying acid-base equilibria, this is a classic example of why pH depends on both concentration and acid strength. Even a solution that seems moderately concentrated can have a relatively mild acidity if its Ka is very small.

Summary: For 0.0555 M boric acid with Ka = 5.8 × 10^-10, the hydrogen ion concentration is approximately 5.67 × 10^-6 M and the pH is approximately 5.25.