Calculate pH of Ammonium Chloride 0.00142 M
Use this interactive calculator to find the pH of an ammonium chloride solution by treating NH4+ as a weak acid. The default example is 0.00142 M NH4Cl at 25 C using the accepted base dissociation constant for ammonia, Kb = 1.8 × 10-5.
For NH4Cl, Cl- is the conjugate base of a strong acid and does not hydrolyze appreciably, so acidity comes from NH4+.
Expert Guide: How to Calculate the pH of 0.00142 M Ammonium Chloride
Ammonium chloride, NH4Cl, is one of the classic examples used to teach acid-base equilibria because it looks neutral at first glance yet actually forms an acidic solution in water. The reason is straightforward: chloride ion is the conjugate base of a strong acid, hydrochloric acid, so it has essentially no effect on pH, while ammonium ion is the conjugate acid of the weak base ammonia and therefore donates protons to water to a small but measurable extent. When the solution concentration is 0.00142 M, the hydrolysis is weak enough that the pH remains only mildly acidic, but strong enough that the difference from neutral pH 7.00 is meaningful and easy to calculate.
If you want the fast answer first, a 0.00142 M ammonium chloride solution at 25 C has a pH of about 6.05 when you use the common value Kb(NH3) = 1.8 × 10-5. That result comes from converting the base dissociation constant of ammonia into the acid dissociation constant of ammonium, then solving the weak acid equilibrium for hydrogen ion concentration. The calculator above performs that work instantly, but understanding the method matters if you are solving chemistry homework, preparing a buffer, checking lab quality control, or comparing concentration effects.
Why ammonium chloride is acidic in water
When NH4Cl dissolves, it dissociates almost completely:
NH4Cl(aq) → NH4+(aq) + Cl–(aq)
The chloride ion is chemically spectator-like for pH purposes in ordinary aqueous solution. The ammonium ion, however, behaves as a weak acid:
NH4+(aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
That equilibrium generates hydronium ions, so the pH drops below 7. The extent of this reaction is governed by the acid dissociation constant of ammonium, Ka. Since many reference tables list Kb for ammonia instead, the usual relationship is:
Ka(NH4+) = Kw / Kb(NH3)
At 25 C, Kw is 1.0 × 10-14. If Kb for ammonia is 1.8 × 10-5, then:
Ka = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
Step-by-step pH calculation for 0.00142 M NH4Cl
Let the initial concentration of ammonium ion be 0.00142 M. Set up an ICE table for the hydrolysis reaction:
- Initial: [NH4+] = 0.00142, [NH3] = 0, [H3O+] = 0
- Change: -x, +x, +x
- Equilibrium: 0.00142 – x, x, x
The equilibrium expression is:
Ka = x2 / (0.00142 – x)
Substitute Ka = 5.56 × 10-10:
5.56 × 10-10 = x2 / (0.00142 – x)
You can solve this exactly with the quadratic formula or approximately using the weak acid assumption that x is much smaller than 0.00142. Because the dissociation is tiny, the approximation is excellent:
x ≈ √(KaC) = √[(5.56 × 10-10)(0.00142)] ≈ 8.88 × 10-7 M
That x value equals [H3O+], so:
pH = -log(8.88 × 10-7) ≈ 6.05
If you solve the quadratic exactly, you obtain essentially the same answer to normal reporting precision. The approximation is justified because x is far below 5 percent of the starting concentration, which is a common check used in general chemistry.
Exact method versus approximation
In weak acid calculations, students often wonder whether the shortcut is acceptable. For 0.00142 M ammonium chloride, the approximation works very well because the acid is weak and the formal concentration is much larger than the amount dissociated. Still, exact treatment is ideal for calculators and formal reports because it removes doubt and remains valid over a wider range of concentrations.
The exact quadratic form comes from rearranging the equilibrium expression into:
x2 + Kax – KaC = 0
Then solve for the physically meaningful positive root:
x = [-Ka + √(Ka2 + 4KaC)] / 2
When C = 0.00142 M and Ka = 5.56 × 10-10, x is again approximately 8.88 × 10-7 M. The pH remains 6.05 to two decimal places.
Accepted constants and reference values
| Quantity | Typical 25 C Value | Why It Matters |
|---|---|---|
| Kw of water | 1.0 × 10-14 | Connects conjugate acid-base pairs and defines neutral water behavior at 25 C. |
| Kb of NH3 | 1.8 × 10-5 | Used to derive Ka for NH4+. |
| Ka of NH4+ | 5.56 × 10-10 | Controls the extent of hydrolysis and final hydrogen ion concentration. |
| pKa of NH4+ | 9.25 | Useful for buffer calculations involving NH3/NH4+. |
How concentration changes the pH of ammonium chloride
The pH of a weakly acidic salt depends strongly on concentration. As ammonium chloride becomes more dilute, the solution becomes less acidic and moves closer to neutral pH. This happens because the hydronium concentration produced by hydrolysis scales roughly with the square root of the formal concentration for a weak acid in the approximation region.
| NH4Cl Concentration (M) | Calculated [H+] | Calculated pH | Interpretation |
|---|---|---|---|
| 0.100 | 7.45 × 10-6 M | 5.13 | Clearly acidic |
| 0.0100 | 2.36 × 10-6 M | 5.63 | Moderately acidic |
| 0.00142 | 8.88 × 10-7 M | 6.05 | Mildly acidic |
| 0.00100 | 7.45 × 10-7 M | 6.13 | Closer to neutral |
| 0.000100 | 2.36 × 10-7 M | 6.63 | Very mildly acidic |
Common mistakes when solving this problem
- Treating NH4Cl as a neutral salt. This is incorrect because NH4+ is the conjugate acid of a weak base.
- Using Kb directly in the acid expression. You must convert Kb of ammonia to Ka of ammonium.
- Forgetting the log step. Once you find [H+], apply pH = -log[H+].
- Ignoring temperature dependence. Kw changes with temperature, so the exact pH can shift outside 25 C conditions.
- Over-rounding intermediate values. Keep enough significant figures to avoid drift in the final pH.
Why the 5 percent rule works here
The weak acid approximation assumes that the amount dissociated, x, is small relative to the initial concentration C. A common benchmark is:
(x / C) × 100% < 5%
For this solution, x ≈ 8.88 × 10-7 M and C = 0.00142 M, so the percent ionization is only about 0.0625 percent. That is far below 5 percent, making the approximation highly reliable. In practical terms, this means the difference between the approximate and exact pH is negligible for routine classwork and quick checks.
Practical laboratory context
Ammonium chloride appears in analytical chemistry, buffer preparation, biological sample handling, and inorganic synthesis. Knowing its pH behavior matters because even weakly acidic salts can influence metal solubility, enzyme stability, reaction rates, and indicator choice. If you prepare a 0.00142 M solution for a calibration or instructional lab, expecting a pH close to 6.05 gives you a rational target. If the measured value is far away from that, possible explanations include contaminated water, dissolved carbon dioxide, pH meter calibration drift, ionic strength effects, or a temperature mismatch.
How to interpret the calculator output
The calculator above reports the concentration, Ka, hydrogen ion concentration, hydroxide concentration, and final pH. It also renders a chart that compares [NH4+] remaining, [H+] generated, and [OH–] present at equilibrium. Because [H+] and [OH–] are much smaller than the formal salt concentration, the graph gives a visual reminder that weak acid hydrolysis alters pH significantly without consuming much of the dissolved ammonium.
Authoritative references for pH and acid-base data
- USGS: pH and Water
- U.S. EPA: pH Overview
- University of Wisconsin: Weak Acids and Equilibrium Concepts
Final answer for 0.00142 M ammonium chloride
Using Kb(NH3) = 1.8 × 10-5 and Kw = 1.0 × 10-14 at 25 C, the conjugate acid constant for NH4+ is 5.56 × 10-10. Solving the hydrolysis equilibrium for a 0.00142 M solution gives [H+] ≈ 8.88 × 10-7 M and therefore pH ≈ 6.05. That makes the solution mildly acidic, exactly as expected for a salt formed from a weak base and a strong acid.