Calculate the pH When 20.0 mL of 0.14 M Stryichine Is Dissolved
Use this interactive calculator to estimate the pH of an aqueous strychnine solution. The tool applies weak base equilibrium chemistry, converts the reported pKa of the conjugate acid into Kb, solves the equilibrium expression, and visualizes the resulting species concentrations.
Calculated Results
Click Calculate pH to solve the weak base equilibrium for strychnine.
How to calculate the pH when 20.0 mL of 0.14 M stryichine is in water
If you searched for how to calculate the pH when 20.0 mL of 0.14 M stryichine is present in solution, the intended compound is almost certainly strychnine, a nitrogen containing organic base. Because strychnine behaves as a weak base in water, you do not treat it the way you would treat sodium hydroxide or another strong base. Instead, you use an equilibrium calculation. That is the key idea behind this page.
At the default settings in the calculator above, the solution pH comes out to about 10.68 at 25 C when the conjugate acid pKa is taken as 8.21. That result makes chemical sense: the solution is definitely basic, but it is not as basic as a strong base of the same formal concentration. Weak bases only partially react with water, so the hydroxide ion concentration is much lower than 0.14 M.
Step 1: Identify the correct acid-base model
Strychnine contains a basic nitrogen that can accept a proton. In water, the base reaction is:
B + H2O ⇌ BH+ + OH-
Here, B is strychnine, BH+ is its conjugate acid, and OH- is the hydroxide produced by hydrolysis. Because the compound is a weak base, the equilibrium constant you need is Kb, not Ka directly.
However, many references list the pKa of the conjugate acid rather than the pKb of the base. That is not a problem. At 25 C:
- Kw = 1.0 x 10^-14
- Ka x Kb = Kw
- Kb = Kw / Ka
Step 2: Convert pKa to Ka and then to Kb
Using the default calculator value pKa = 8.21 for the conjugate acid of strychnine:
- Find Ka: Ka = 10^-8.21 = 6.17 x 10^-9
- Find Kb: Kb = (1.0 x 10^-14) / (6.17 x 10^-9) = 1.62 x 10^-6
That Kb value tells you strychnine is a weak base, but not an extremely weak one. It hydrolyzes enough to generate measurable hydroxide and push pH well above neutral.
| Quantity | Value | Meaning in this problem |
|---|---|---|
| Volume | 20.0 mL = 0.0200 L | Needed to compute total moles present |
| Formal concentration | 0.14 M | Starting base concentration in solution |
| Moles of strychnine | 0.14 x 0.0200 = 0.00280 mol | Total amount of base present |
| Conjugate acid pKa | 8.21 | Used to derive Ka and Kb |
| Ka of BH+ | 6.17 x 10^-9 | Acid strength of protonated strychnine |
| Kb of B | 1.62 x 10^-6 | Base strength of strychnine |
Step 3: Set up the ICE table
For the equilibrium
B + H2O ⇌ BH+ + OH-
start with:
- [B]initial = 0.14 M
- [BH+]initial = 0
- [OH-]initial ≈ 0
Let x be the amount of base that reacts:
- [B]eq = 0.14 – x
- [BH+]eq = x
- [OH-]eq = x
Now write the equilibrium expression:
Kb = [BH+][OH-] / [B] = x² / (0.14 – x)
Step 4: Solve for x, which equals [OH-]
Substitute the numbers:
1.62 x 10^-6 = x² / (0.14 – x)
Rearranging gives the quadratic form:
x² + Kb x – KbC = 0
where C = 0.14.
The exact solution is:
x = (-Kb + √(Kb² + 4KbC)) / 2
Substituting the values gives:
x ≈ 4.75 x 10^-4 M
Because x represents the hydroxide concentration generated by the weak base reaction, we now have:
- [OH-] = 4.75 x 10^-4 M
- pOH = -log(4.75 x 10^-4) ≈ 3.32
- pH = 14.00 – 3.32 = 10.68
Why the 20.0 mL volume matters less than many students expect
One of the most common points of confusion in this type of problem is the role of volume. Students often see “20.0 mL of 0.14 M strychnine” and assume the volume must directly affect pH. In this case, if the concentration is already given and there is no later dilution or titration step, the pH depends on:
- the formal concentration of the weak base
- the base strength expressed through Kb
- the temperature assumption, usually 25 C for textbook problems
The volume is still useful because it tells you the number of moles present:
moles = M x L = 0.14 x 0.0200 = 0.00280 mol
But if those 0.00280 mol are contained in 20.0 mL, the concentration is already 0.14 M, and that is the concentration that controls the equilibrium position.
Approximation method versus exact quadratic method
In weak acid and weak base problems, many instructors allow the square root approximation if the extent of ionization is small. The approximation assumes 0.14 – x ≈ 0.14, so:
x ≈ √(KbC) = √((1.62 x 10^-6)(0.14)) ≈ 4.76 x 10^-4 M
This is almost identical to the exact answer because the percent ionization is low:
(4.75 x 10^-4 / 0.14) x 100 ≈ 0.34%
Since the ionization is much less than 5%, the approximation is excellent.
| Method | [OH-] (M) | pOH | pH | Comment |
|---|---|---|---|---|
| Exact quadratic | 4.75 x 10^-4 | 3.32 | 10.68 | Most rigorous classroom answer |
| Square root approximation | 4.76 x 10^-4 | 3.32 | 10.68 | Valid because ionization is only about 0.34% |
| Incorrect strong base assumption | 0.14 | 0.85 | 13.15 | Far too high because strychnine is not a strong base |
What real chemical data tell you about this result
Real acid-base constants matter. For weak bases, a small shift in pKa can change the final pH by a few hundredths to a few tenths of a pH unit. That is why the calculator lets you adjust the pKa value. Published values for alkaloids can vary slightly depending on the source, ionic strength, solvent conditions, and temperature. Even so, the pH for a 0.14 M strychnine solution remains clearly basic.
Sensitivity of pH to the chosen pKa value
The table below shows how the final pH changes when the conjugate acid pKa changes over a realistic narrow range for a weak organic base. These values are calculated at 25 C for a 0.14 M solution.
| Assumed pKa of BH+ | Derived Kb | Calculated [OH-] (M) | Calculated pH |
|---|---|---|---|
| 8.00 | 1.00 x 10^-6 | 3.74 x 10^-4 | 10.57 |
| 8.10 | 1.26 x 10^-6 | 4.20 x 10^-4 | 10.62 |
| 8.21 | 1.62 x 10^-6 | 4.75 x 10^-4 | 10.68 |
| 8.30 | 2.00 x 10^-6 | 5.28 x 10^-4 | 10.72 |
Common mistakes to avoid
- Treating strychnine as a strong base. If you do that, you massively overestimate the hydroxide concentration and pH.
- Using the pKa directly as if it were pKb. The given pKa applies to the conjugate acid, so you must convert it first.
- Ignoring the equilibrium setup. You need an ICE table or an equivalent equilibrium expression to solve weak base problems correctly.
- Forgetting the pOH to pH conversion. Once you calculate [OH-], find pOH first, then use pH = 14.00 – pOH at 25 C.
- Overusing volume. In a simple fixed-concentration sample, volume gives moles but does not directly change pH unless dilution or reaction with another solution occurs.
When this calculator is most useful
This page is ideal when you need to:
- check a homework or quiz style weak base pH problem
- compare exact and approximation based methods
- see how the pKa assumption changes final pH
- understand the relation between concentration, ionization, and pH
- visualize the concentration of unprotonated base, protonated base, and hydroxide at equilibrium
Authoritative chemistry and pH references
For additional background on pH, aqueous chemistry, and chemical constants, these sources are useful starting points:
Final takeaway
To calculate the pH when 20.0 mL of 0.14 M stryichine, meaning strychnine, is dissolved in water, treat the compound as a weak base. Convert the conjugate acid pKa to Ka, then to Kb, write the weak base equilibrium, solve for hydroxide concentration, and convert that to pOH and pH. Using a conjugate acid pKa of 8.21 at 25 C gives:
- moles of strychnine = 0.00280 mol
- Kb = 1.62 x 10^-6
- [OH-] ≈ 4.75 x 10^-4 M
- pOH ≈ 3.32
- pH ≈ 10.68
That is the chemically correct weak base answer, and it aligns with what the calculator above will return using its default values.