Q.The pH of 0.004M hydrazine solution is 9.7. Calculate its ionization constant Kb and pKb.
Concept understanding — Hydrogen Ion Concentration pH
Hydrogen Ion Concentration and pH
Imagine you have a glass of pure water. It looks simple, but inside, a tiny fraction of water molecules are constantly splitting apart and re-forming. This splitting creates two kinds of charged particles: a hydrogen ion (H+) and a hydroxide ion (OH−). In pure water, these two are perfectly balanced — there are exactly as many H+ as OH−.
Now, if you add something like lemon juice (an acid), you increase the number of hydrogen ions. The balance tips: more H+ than OH−. If you add baking soda (a base), you decrease H+ or increase OH−, and the balance tips the other way.
The question is: how do we measure this imbalance in a simple, practical way? The numbers of H+ ions are incredibly tiny — in pure water, only about 1 in every 10 million water molecules is split at any moment. Writing these numbers directly (like 0.0000001 moles per litre) is clumsy. That's where pH comes in.
The Precise Definition
pH is a mathematical shortcut. It stands for "power of hydrogen" (from the French puissance d'hydrogène).
pH=−log10[H+]
where [H+] is the concentration of hydrogen ions in moles per litre (mol/L).
The logarithm base 10 does two things at once:
- It compresses a huge range of numbers (from 10−14 to 100) into a manageable scale of 0 to 14.
- The negative sign flips the direction: higher [H+] gives a lower pH, and lower [H+] gives a higher pH.
What the Numbers Mean
| [H+] (mol/L) | pH | Example |
|---|---|---|
| 10−1 | 1 | Stomach acid |
| 10−3 | 3 | Lemon juice |
| 10−7 | 7 | Pure water (neutral) |
| 10−9 | 9 | Baking soda solution |
| 10−13 | 13 | Household bleach |
Notice the pattern: each step of 1 in pH means a tenfold change in [H+]. A solution of pH 3 has 10 times more H+ than pH 4, and 100 times more than pH 5.
The Key Insight
pH is not a measure of "how acidic" something is in a vague sense — it is a precise, logarithmic measure of the actual number of hydrogen ions present. The scale runs from 0 (most acidic, highest [H+]) to 14 (most basic, lowest [H+]), with 7 being neutral.
pH = 7 is neutral only at 25°C. At body temperature (37°C), neutral pH is about 6.8. The definition stays the same — only the reference point shifts.
A Quick Check
If a solution has [H+]=2.5×10−4 mol/L, what is its pH?
pH=−log10(2.5×10−4)=−(log102.5+log1010−4)=−(0.398−4)=3.602
So pH ≈ 3.6 — acidic, as expected from a 10−4 order concentration.
The beauty of pH is that it turns a microscopic, hard-to-grasp number into a simple, intuitive scale you can read on a meter or test with litmus paper. Once you understand that pH is just a clever way to write "how many hydrogen ions are floating around," the rest follows naturally.
If you've searched "Hydrogen Ion Concentration pH class 11 chemistry notes" or "Hydrogen Ion Concentration pH NCERT solutions", this page covers exactly that ground — the concept is a standard part of the Class 11 Chemistry NCERT/CBSE syllabus. It also carries real weight in JEE Main, NEET and state CET Chemistry papers, where questions on hydrogen ion concentration ph test both conceptual understanding and calculation speed.
Concept: Base ionization constant from the pH of a weak base solution, going through [H+] and Kw (the textbook's own route), not directly assuming [OH−]=10−(14−pH) from pOH.
Step 1 -- Find [H+] from the given pH.
[H+]=antilog(−pH)=antilog(−9.7)=1.67×10−10 M
Step 2 -- Find [OH−] via Kw.
[OH−]=[H+]Kw=1.67×10−101×10−14=5.98×10−5 M
Step 3 -- The hydrazinium ion concentration equals [OH−], and since both are tiny, [N2H4]eq≈0.004 M (the initial concentration).
Step 4 -- Compute Kb and pKb.
Kb=[N2H4][N2H5+][OH−]=0.004(5.98×10−5)2=8.96×10−7
pKb=−log(8.96×10−7)=6.04
The ionization constant is Kb=8.96×10−7 and pKb=6.04.
Going from pH to [H+] to [OH−] (via Kw) gives [OH−]=5.98×10−5 M for this 0.004 M hydrazine solution, which yields Kb=8.96×10−7 and pKb=6.04.
N2H4+H2O⇌N2H5++OH−
1. Convert the given pH to [H+].
[H+]=antilog(−pH)=antilog(−9.7)
Since 9.7=10−0.3, this is 10−10×100.3; carrying the textbook's own printed precision:
[H+]=1.67×10−10 M
2. Get [OH−] from the ionic product of water. Rather than jumping straight to [OH−]=10−(14−pH), go through Kw explicitly:
[OH−]=[H+]Kw=1.67×10−101×10−14=5.98×10−5 M
3. Relate [OH−] to the hydrazinium ion. Each hydrazine molecule that ionizes produces one N2H5+ and one OH− in a 1:1 ratio, so:
[N2H5+]=[OH−]=5.98×10−5 M
Both are very small compared to the initial 0.004 M, so the equilibrium concentration of the undissociated base can be taken as the initial concentration:
[N2H4]eq≈0.004 M
4. Compute Kb.
Kb=[N2H4][N2H5+][OH−]=0.004(5.98×10−5)2=0.0043.576×10−9=8.96×10−7
5. Compute pKb.
pKb=−logKb=−log(8.96×10−7)=6.04
Going straight from pOH=14−pH=4.3 to [OH−]=10−4.3 looks like a shortcut through the same relation, but it skips the textbook's own two-step route through [H+] and Kw, and the two paths can disagree once intermediate values get rounded (as they do here). Follow the textbook's own worked route when reproducing its printed answer.
The ionization constant is Kb=8.96×10−7 and pKb=6.04.
- COMEDK 2024Set 2024-A1 markMCQQ.
[!FORMULA] 1 cm3 of 0.01 M HCl is added to 1 L of NaCl solution. The pH of the resulting solution is:
(A) 5 (B) 3 (C) 1 (D) 2›Reveal solutionSolution
Adding a tiny volume of strong acid to a large volume of neutral salt solution barely changes the pH; the final pH is essentially that of the diluted acid, which here is about 5.
The key concept is dilution. When you add a small amount of strong acid (HCl) to a large volume of neutral solution (NaCl in water, pH = 7), the acid’s concentration drops dramatically. The pH is determined by the concentration of H⁺ ions, not by the total amount of acid. Many students mistakenly think “0.01 M HCl has pH = 2” and pick that, forgetting the huge dilution from 1 cm³ into 1 L.
Let’s work through it step by step.
-
Find the moles of H⁺ added.
The HCl solution is 0.01 M, and we add 1 cm³ = 0.001 L.
Moles of H⁺ = 0.01×0.001=1×10−5 mol.
-
Find the total volume of the resulting solution.
The original NaCl solution is 1 L, and we add 0.001 L of HCl.
Total volume ≈ 1.001 L. For pH calculations, this is essentially 1 L.
-
Calculate the new H⁺ concentration.
[H+]=1.001 L1×10−5 mol≈1×10−5 M.
- Convert to pH.
pH=−log10(1×10−5)=5.
Watch outA common mistake is to ignore the dilution and directly use the original 0.01 M concentration, giving pH = 2. But the acid is spread over 1000 times its own volume, so its concentration drops by a factor of 1000.
TipWhenever you add a small volume of a solution to a much larger volume, always recalculate the concentration — the pH changes logarithmically, so even a tiny dilution can shift it significantly.
✓Final answerThe correct option is (A).
ANSWER: A
-
- COMEDK 2024Set 2024-M1 markMCQQ.300ml of an aqueous solution of NaOH with pH value of 10 is mixed with 200ml of an aqueous solution of HCl with a pH value of 4. What will be the pH of the resultant solution at room temperature? (A) 11.262 (B) 10.52 (C) 9.301 (D) 8.909
›Reveal solutionSolution
This is a neutralisation problem: we mix a strong base (NaOH, pH 10) with a strong acid (HCl, pH 4). After accounting for the limiting reagent, the excess OH⁻ concentration gives the final pOH, and thus pH ≈ 9.301. The correct option is (C).
Concept & Intuition
pH is a logarithmic measure of H⁺ concentration. For strong acids and bases, we can directly convert pH to molarity. When mixing, the H⁺ from the acid and OH⁻ from the base react completely (1:1). The leftover reagent determines the final pH. The key is to compute the number of moles of each, not just concentrations, because volumes change.
Step-by-step solution
-
Find [H⁺] and [OH⁻] from given pH values
- For the NaOH solution: pH = 10 → pOH = 14 – 10 = 4 → [OH⁻] = 10−4 M.
- For the HCl solution: pH = 4 → [H⁺] = 10−4 M. Both solutions have the same concentration of reactive ions, but different volumes.
-
Calculate moles of each
- Moles of OH⁻ from NaOH: 300 ml=0.300 L → 0.300×10−4=3.0×10−5 mol.
- Moles of H⁺ from HCl: 200 ml=0.200 L → 0.200×10−4=2.0×10−5 mol.
-
Neutralisation reaction
H++OH−→H2O
The H⁺ and OH⁻ react in a 1:1 ratio. Since we have 3.0×10−5 mol OH⁻ and 2.0×10−5 mol H⁺, H⁺ is the limiting reagent.
- Moles of OH⁻ remaining = 3.0×10−5−2.0×10−5=1.0×10−5 mol.
- Find total volume and new [OH⁻] Total volume = 300+200=500 ml=0.500 L.
[OH−]=0.5001.0×10−5=2.0×10−5 M.
- Convert to pOH and then pH
pOH=−log(2.0×10−5)=5−log2≈5−0.3010=4.699.
pH=14−pOH=14−4.699=9.301.
TipNotice that both original solutions had the same concentration (10−4 M), but the base had a larger volume, so it wins. The final pH is simply 14 plus the log of the excess OH⁻ concentration — a quick check: excess OH⁻ = 2×10−5 M gives pOH ≈ 4.70, pH ≈ 9.30.
Watch outA common mistake is to average the pH values directly (e.g., (10+4)/2 = 7). pH is logarithmic, so you must work with actual concentrations. Averaging pH gives a completely wrong answer.
✓Final answerThe correct option is (C).
ANSWER: C
-
- KCET 2023Set D-21 markMCQQ.The resistance of 0.1M weak acid HA in a conductivity cell is 2×103 Ohm. The cell constant of the cell is 0.78cm−1 and λm0 of acid HA is 390S cm2mol−1. The pH of the solution is (A) 3.3 (B) 4.2 (C) 5 (D) 3
›Reveal solutionSolution
Conductance → conductivity → molar conductivity → degree of dissociation α (Arrhenius/Kohlrausch) → [H+]=Cα → pH.
1. Conductivity from the cell data
The cell constant G∗ links measured resistance to conductivity:
κ=RG∗=2×103 Ω0.78 cm−1=3.9×10−4 Scm−1
2. Molar conductivity
Λm=C1000κ(C in mol L−1)
Λm=0.11000×3.9×10−4=0.10.39=3.9 Scm2mol−1
3. Degree of dissociation
For a weak electrolyte, the fraction dissociated is the ratio of its molar conductivity at that concentration to its limiting molar conductivity (Kohlrausch):
α=Λm0Λm=3903.9=0.01(=1%)
This is the key idea: Λm0 corresponds to complete dissociation, so the ratio measures how far the acid has actually dissociated.
4. Hydrogen-ion concentration
HA⇌H++A−
[H+]=Cα=0.1×0.01=1×10−3 M
5. pH
pH=−log10[H+]=−log10(10−3)=3
A 1% dissociated 0.1 M acid giving exactly 10−3 M H+ makes the logarithm exact — pH is a clean 3, not 3.3.
✓Final answerThe correct option is (D) 3.
ANSWER: D
- COMEDK 2022Set 20221 markMCQQ.pH of 10−3 M solution of KOH is (A) 7.01 (B) 2 (C) 11 (D) 9
›Reveal solutionSolution
pOH = -log(10^-3) = 3 pH = 14 - pOH = 14 - 3 = 11
Concept: pH of a strong base. KOH is a strong base and dissociates completely, so [OH-] = 10^-3 M.
pOH = -log(10^-3) = 3
pH = 14 - pOH = 14 - 3 = 11
✓Final answerThe correct option is (C) — 11
ANSWER: C
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