Calculate The Ph Of A 0.64 M Solution Of C2H5Nh3Cl

Calculate the pH of a 0.64 M Solution of C2H5NH3Cl

Use this premium acid-base calculator to determine the pH of ethylammonium chloride solution, review the chemistry behind the calculation, and visualize hydronium, hydroxide, and conjugate-acid concentrations with an interactive chart.

Ethylammonium Chloride pH Calculator

pH = 5.47
Ka of C2H5NH3+ 1.79 × 10-11
[H3O+] 3.39 × 10-6 M
[OH-] 2.95 × 10-9 M
Acid behavior Weakly acidic aqueous salt
For a 0.64 M solution of C2H5NH3Cl, using Kb = 5.6 × 10-4 for ethylamine gives Ka = Kw/Kb = 1.79 × 10-11. Solving for hydronium concentration gives a pH of about 5.47 at 25°C.

Species Concentration Profile

Chart comparing the dominant dissolved species after hydrolysis of ethylammonium ion in water.

Default salt C2H5NH3Cl
Conjugate acid C2H5NH3+
Expected solution type Mildly acidic

How to Calculate the pH of a 0.64 M Solution of C2H5NH3Cl

To calculate the pH of a 0.64 M solution of C2H5NH3Cl, you need to recognize that this compound is ethylammonium chloride, a salt formed from a weak base and a strong acid. The chloride ion, Cl, comes from hydrochloric acid and is essentially pH-neutral in water. The chemically important ion is the ethylammonium cation, C2H5NH3+, which is the conjugate acid of the weak base ethylamine, C2H5NH2.

That means the solution is not neutral. Instead, the cation undergoes hydrolysis in water:

C2H5NH3+ + H2O ⇌ C2H5NH2 + H3O+

Because hydronium ions are produced, the solution becomes acidic. The pH is determined by the acid dissociation constant of C2H5NH3+, not by simply assuming full dissociation like a strong acid. This distinction is the heart of the problem and is the reason many students initially overestimate the acidity.

Step 1: Identify the Relevant Equilibrium Constant

Most tables list Kb for ethylamine rather than Ka for ethylammonium. A common value for ethylamine at 25°C is:

Kb(C2H5NH2) = 5.6 × 10^-4

Since the conjugate acid-base pair is related by:

Ka × Kb = Kw

and at 25°C:

Kw = 1.0 × 10^-14

you can solve for Ka:

Ka = Kw / Kb = (1.0 × 10^-14) / (5.6 × 10^-4) = 1.79 × 10^-11

This small Ka value tells you that C2H5NH3+ is a weak acid, so the pH should be below 7 but not dramatically low.

Step 2: Set Up the ICE Table

For a 0.64 M solution, the initial concentration of C2H5NH3+ is 0.64 M. Let x represent the amount that dissociates.

C2H5NH3+ + H2O ⇌ C2H5NH2 + H3O+
  • Initial: [C2H5NH3+] = 0.64, [C2H5NH2] = 0, [H3O+] = 0
  • Change: -x, +x, +x
  • Equilibrium: 0.64 – x, x, x

Substitute into the acid expression:

Ka = x^2 / (0.64 – x) = 1.79 × 10^-11

Step 3: Use the Weak Acid Approximation

Because Ka is tiny, x will be much smaller than 0.64, so you can approximate:

0.64 – x ≈ 0.64

Then:

x^2 / 0.64 = 1.79 × 10^-11
x^2 = 1.1456 × 10^-11
x = [H3O+] = 3.39 × 10^-6 M

Now calculate the pH:

pH = -log(3.39 × 10^-6) = 5.47
Final answer: the pH of a 0.64 M solution of C2H5NH3Cl is approximately 5.47 at 25°C when Kb for ethylamine is taken as 5.6 × 10-4.

Why C2H5NH3Cl Is Acidic Instead of Neutral

This salt contains two ions after dissolving:

  • C2H5NH3+, the conjugate acid of a weak base
  • Cl, the conjugate base of a strong acid

Chloride does not significantly react with water, but the ethylammonium ion donates protons weakly to water. This makes the solution acidic. A useful rule is:

  • Strong acid + strong base salt: usually neutral
  • Strong acid + weak base salt: acidic
  • Weak acid + strong base salt: basic

Ethylammonium chloride falls squarely into the second category. That is why a student who memorizes only “salts dissociate completely” may miss the hydrolysis step and get the wrong pH.

Exact Solution Versus Approximation

The weak-acid approximation is excellent here, but it is good chemistry practice to know why. The exact quadratic form starts from:

Ka = x^2 / (0.64 – x)

Rearrange:

x^2 + Ka x – Ka(0.64) = 0

Substituting Ka = 1.79 × 10-11 gives an exact hydronium concentration essentially identical to the approximation. The percent ionization is tiny:

Percent ionization = (x / 0.64) × 100 ≈ 0.00053%

Since this is far below 5%, the approximation is fully justified. In classroom settings, this is often all you need to show to validate using the square-root shortcut.

Comparison Table: Similar Weak Base Conjugate Acids

The acidity of ammonium-like ions depends on the basicity of the parent amine. Stronger parent bases produce weaker conjugate acids. The table below compares a few related systems using representative 25°C values.

Conjugate acid Parent base Representative Kb of base Calculated Ka of conjugate acid Acid strength trend
NH4+ NH3 1.8 × 10-5 5.6 × 10-10 Stronger acid than ethylammonium
CH3NH3+ CH3NH2 4.4 × 10-4 2.3 × 10-11 Similar weak acidity
C2H5NH3+ C2H5NH2 5.6 × 10-4 1.79 × 10-11 Weak acid

The key takeaway is that ethylamine is a stronger base than ammonia, so ethylammonium is a weaker acid than ammonium. That is one reason the pH of ethylammonium chloride solution is not nearly as low as a similarly concentrated strong acid solution.

How Concentration Affects the pH

For weak acids, pH depends on both Ka and concentration. If the concentration of C2H5NH3Cl changes, the pH changes too. Under the weak-acid approximation, hydronium concentration is proportional to the square root of Ka times concentration:

[H3O+] ≈ √(KaC)

This means a tenfold increase in concentration does not decrease the pH by a full 1.00 unit as it would for a strong monoprotic acid. Instead, the change is more gradual.

C2H5NH3Cl concentration (M) Approximate [H3O+] (M) Approximate pH Interpretation
0.010 4.23 × 10-7 6.37 Only slightly acidic
0.10 1.34 × 10-6 5.87 Mildly acidic
0.64 3.39 × 10-6 5.47 Moderately more acidic, still weak acid behavior
1.00 4.23 × 10-6 5.37 Acidity rises slowly with concentration

Common Mistakes Students Make

  1. Treating C2H5NH3Cl like a strong acid. The salt dissolves completely, but the acid-base behavior comes from hydrolysis of the conjugate acid, not from direct strong-acid dissociation.
  2. Using Kb directly in the acid calculation. You must first convert Kb of ethylamine into Ka for ethylammonium using Ka = Kw/Kb.
  3. Ignoring chloride’s role correctly. Chloride is a spectator ion in this context. It does not make the solution acidic.
  4. Forgetting the temperature assumption. Most textbook pH calculations assume 25°C, where Kw = 1.0 × 10-14.
  5. Not checking the approximation. Even though the approximation works here, it is still proper technique to compare x to the initial concentration.

Practical Interpretation of the Result

A pH of 5.47 indicates a solution that is acidic, but only mildly so. It is far less acidic than a 0.64 M strong acid solution, which would have a pH close to 0.19. This enormous difference highlights how much acid strength matters. Concentration alone does not determine pH; the equilibrium constant is equally important.

In laboratory practice, salts of protonated amines are common in synthesis, pharmaceutical chemistry, and analytical chemistry because they are often more stable or more water-soluble than the corresponding free amines. Understanding the pH of these solutions can matter for:

  • buffer design and pH adjustment
  • solubility control
  • reaction selectivity
  • storage and formulation conditions

Authoritative Chemistry References

If you want to verify acid-base fundamentals, equilibrium relationships, or broader water chemistry concepts, these sources are useful:

Worked Summary

Here is the full solution in compact form:

  1. Recognize C2H5NH3Cl as a salt of weak base + strong acid.
  2. Only C2H5NH3+ hydrolyzes significantly.
  3. Use Kb(C2H5NH2) = 5.6 × 10-4.
  4. Compute Ka = 1.0 × 10-14 / 5.6 × 10-4 = 1.79 × 10-11.
  5. Set up Ka = x2 / (0.64 – x).
  6. Approximate x small, so x ≈ √(Ka × 0.64) = 3.39 × 10-6.
  7. Compute pH = -log(3.39 × 10-6) = 5.47.
Bottom line: the pH of a 0.64 M solution of C2H5NH3Cl is about 5.47, assuming 25°C and Kb = 5.6 × 10-4 for ethylamine.

FAQ: Quick Answers About This Calculation

Is C2H5NH3Cl a strong acid?

No. It is a salt. Its acidity comes from the weakly acidic conjugate acid C2H5NH3+ reacting with water.

Why not use pH = -log(0.64)?

That shortcut applies to a strong acid that fully generates hydronium equal to its formal concentration. Ethylammonium chloride does not do that.

Does chloride affect the pH?

In this problem, chloride is essentially a spectator ion because it is the conjugate base of a strong acid.

Can I use the approximation every time?

Not always. It works here because the percent ionization is extremely small. In more concentrated or stronger weak-acid systems, you should verify the assumption or solve the quadratic exactly.

This calculator uses standard aqueous equilibrium assumptions at 25°C. Actual laboratory pH can vary slightly with ionic strength, activity corrections, and source-dependent literature values of Kb.

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