Q.The pH of 0.004M hydrazine solution is 9.7. Calculate its ionization constant Kb and pKb.

Showing the 12 most recent of 20 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.The pH of a solution obtained by mixing 250 mL of 0.5 M NaOH, 100 mL of 0.5 M HCl and 400 mL of water is (A) 12 (B) 2 (C) 1 (D) 13

   ›Reveal solutionSolution

   A strong-base/strong-acid mixing problem: find the millimoles of leftover strong base after neutralization, divide by total volume, and convert to pH. Answer: pH $=13$.

   Concept and Intuition

   When a strong acid and a strong base are mixed, $\mathrm{H^+}$ and $\mathrm{OH^-}$ react quantitatively (1:1) until the limiting reagent is exhausted. Whatever species is in excess — acid or base — sets the final $\mathrm{pH}$ of the diluted mixture; water added afterward is a pure diluent and does not add or remove moles of acid/base, it just increases the volume over which the leftover ions are spread.

   Step-by-Step Solution

   1. Millimoles of NaOH $= 250\ \mathrm{mL} \times 0.5\ \mathrm{mol/L} = 125\ \mathrm{mmol}$.
   2. Millimoles of HCl $= 100\ \mathrm{mL} \times 0.5\ \mathrm{mol/L} = 50\ \mathrm{mmol}$.
   3. $\mathrm{H^+}$ and $\mathrm{OH^-}$ neutralize 1:1: leftover $\mathrm{OH^-} = 125-50 = 75\ \mathrm{mmol}$ (base is in excess).
   4. Total volume of the final mixture $= 250+100+400 = 750\ \mathrm{mL}$ (the water dilutes but adds no moles).
   5. $[\mathrm{OH^-}] = \dfrac{75\ \mathrm{mmol}}{750\ \mathrm{mL}} = 0.1\ \mathrm{mol/L}$.
   6. $\mathrm{pOH} = -\log(0.1) = 1$, so $\mathrm{pH} = 14 - 1 = 13$.

   Common Mistakes

   • Forgetting to add the 400 mL of water into the total volume used for dilution.
   • Computing pH directly from the leftover moles of $\mathrm{OH^-}$ without converting to $\mathrm{pOH}$ first (i.e., reporting pH $=1$ instead of $13$).
   • Mixing up which reagent is in excess (checking $125>50$ confirms it's the base, not the acid).

   ✓Final answer

   The correct option is (D) — pH = 13.

   ANSWER: D

2. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.The pH of 1 L of HCl solution is 1.0. What is the volume (in L) of water to be added to this solution to increase its pH to 2.0? (A) 10 (B) 9 (C) 100 (D) 99

   ›Reveal solutionSolution

   Dilution of a strong acid raises pH by increasing volume while keeping moles of $H^+$ fixed; here 9 L of water must be added.

   Concept and Intuition

   HCl is a strong acid, so it dissociates completely: $[H^+]$ equals the acid concentration. pH depends only on $[H^+]$, and diluting the solution lowers $[H^+]$ without destroying any $H^+$ ions — the moles of $H^+$ stay constant; only the volume (and hence concentration) changes.

   Step-by-Step Solution

   1. At pH $=1.0$: $[H^+] = 10^{-1} = 0.1\ M$. In 1 L, moles of $H^+ = 0.1 \times 1 = 0.1$ mol.
   2. Target pH $=2.0$: $[H^+] = 10^{-2} = 0.01\ M$.
   3. Moles of $H^+$ are conserved on dilution: $0.1\ \text{mol} = 0.01\ M \times V_{final}$.
   4. $V_{final} = \dfrac{0.1}{0.01} = 10\ L$.
   5. Water added $= V_{final} - V_{initial} = 10 - 1 = 9\ L$.

   Common Mistakes

   • Forgetting to subtract the original 1 L and reporting the final volume (10 L) instead of the water added (9 L).
   • Treating $[H^+]$ itself (not moles) as conserved during dilution.

   ✓Final answer

   The correct option is (B) — 9.

   ANSWER: B

3. AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
   Q.Which of the following aqueous solutions has highest pH ? (Given $\log 2 = 0.30$, $\log 3 = 0.48$, $\log 4 = 0.60$, $\log 5 = 0.70$) (A) $0.2\ M\ \mathrm{Ba(OH)_2}$ (B) $0.02\ N\ \mathrm{Ba(OH)_2}$ (C) $0.1\ M\ NaOH$ (D) $0.05\ M\ \mathrm{Ba(OH)_2}$

   ›Reveal solutionSolution

   Converting every option to $[OH^-]$ shows 0.2 M $\mathrm{Ba(OH)_2}$ gives the largest hydroxide concentration (0.4 M), hence the highest pH.

   Concept and Intuition

   pH is highest when $[OH^-]$ (for a basic solution) is highest, since $\text{pOH} = -\log[OH^-]$ and $\text{pH}=14-\text{pOH}$. Each base here dissociates completely (strong base), so $[OH^-]$ is simply the molarity times the number of $OH^-$ ions per formula unit — take care converting normality to molarity where needed.

   Step-by-Step Solution

   1. (A) 0.2 M $\mathrm{Ba(OH)_2}$: each formula unit gives 2 $OH^-$, so $[OH^-] = 2\times0.2 = 0.4$ M.
   2. (B) 0.02 N $\mathrm{Ba(OH)_2}$: n-factor for $\mathrm{Ba(OH)_2}$ is 2 (2 replaceable $OH^-$), so molarity $= 0.02/2 = 0.01$ M, giving $[OH^-] = 2\times0.01 = 0.02$ M.
   3. (C) 0.1 M NaOH: 1 $OH^-$ per formula unit, so $[OH^-] = 0.1$ M.
   4. (D) 0.05 M $\mathrm{Ba(OH)_2}$: $[OH^-] = 2\times0.05 = 0.1$ M.
   5. Comparing: (A) 0.4 M > (C)=(D) 0.1 M > (B) 0.02 M. So (A) has the highest $[OH^-]$, hence the highest pH.

   Common Mistakes

   • Forgetting that $\mathrm{Ba(OH)_2}$ releases two hydroxide ions per formula unit, not one.
   • Confusing normality with molarity for a base with n-factor 2 — normality is twice the molarity for $\mathrm{Ba(OH)_2}$, so molarity is normality divided by 2.

   ✓Final answer

   The correct option is (A) — 0.2 M $\mathrm{Ba(OH)_2}$.

   ANSWER: A

4. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.At $27^{\circ}C$, 100 mL of 0.4 M HCl is mixed with 100 mL of 0.5 M NaOH solution. To the resultant solution, 800 mL of distilled water is added. What is the pH of final solution? (A) 12 (B) 2 (C) 1.3 (D) 1.0

   ›Reveal solutionSolution

   Excess NaOH after neutralization, diluted to 1 L total, gives 0.01 M OH⁻ ⇒ pH = 12 — answer (A).

   Concept and Intuition

   When a strong acid and a strong base are mixed, they neutralize mole-for-mole; whichever reagent is in excess determines whether the final solution is acidic or basic. Diluting afterward simply divides the moles of the leftover species by the new total volume to get its final concentration.

   Step-by-Step Solution

   1. Moles of HCl $= 0.100\ \mathrm L \times 0.4\ \mathrm{mol/L} = 0.04\ \mathrm{mol}$.
   2. Moles of NaOH $= 0.100\ \mathrm L \times 0.5\ \mathrm{mol/L} = 0.05\ \mathrm{mol}$.
   3. $\mathrm{HCl + NaOH \to NaCl + H_2O}$ consumes 0.04 mol of each; excess NaOH left $= 0.05-0.04 = 0.01\ \mathrm{mol}$.
   4. Total volume after adding 800 mL water $= 100+100+800 = 1000\ \mathrm{mL} = 1\ \mathrm L$.
   5. $[\mathrm{OH^-}] = 0.01\ \mathrm{mol}/1\ \mathrm L = 0.01\ \mathrm M = 10^{-2}\ \mathrm M$.
   6. $\mathrm{pOH} = -\log(10^{-2}) = 2$; $\mathrm{pH} = 14 - 2 = 12$.

   Common Mistakes

   • Forgetting to add the 800 mL water into the total volume (using 200 mL instead of 1000 mL).
   • Mixing up pH and pOH at the last step.

   ✓Final answer

   The correct option is (A) — 12.

   ANSWER: A

5. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
   Q.Which of the following when added to 20 mL of a 0.01 M solution of HCl would decrease its pH? (A) 20 mL of 0.02 M HCl (B) 20 mL of 0.005 M HCl (C) 20 mL of 0.01 M HCl (D) 40 mL of 0.005 M HCl

   ›Reveal solutionSolution

   Mixing HCl solutions changes concentration depending on whether the added solution is more or less concentrated than the original; only adding a more concentrated HCl (20 mL of 0.02 M) raises [H⁺] and lowers pH.

   Concept and Intuition

   pH decreases when [H⁺] increases. Mixing two HCl solutions gives a new concentration equal to (total moles of H⁺)/(total volume). If the added solution is more concentrated than the original, the mixture's concentration rises above the original, decreasing pH; if less concentrated (or equal, but diluted with more solvent), the mixture's concentration falls or stays the same.

   Step-by-Step Solution

   1. Original: 20 mL of 0.01 M HCl → 0.2 mmol H⁺ in 20 mL.
   2. (A) Add 20 mL of 0.02 M HCl: total H⁺ $=0.2+0.4=0.6$ mmol in $40$ mL $\Rightarrow 0.015$ M $>0.01$ M — concentration increases, pH decreases. ✓
   3. (B) Add 20 mL of 0.005 M HCl: total H⁺ $=0.2+0.1=0.3$ mmol in 40 mL $\Rightarrow 0.0075$ M $<0.01$ M — pH increases.
   4. (C) Add 20 mL of 0.01 M HCl (same concentration): total H⁺ $=0.2+0.2=0.4$ mmol in 40 mL $\Rightarrow 0.01$ M — unchanged, pH unchanged.
   5. (D) Add 40 mL of 0.005 M HCl: total H⁺ $=0.2+0.2=0.4$ mmol in 60 mL $\Rightarrow 0.00667$ M $<0.01$ M — pH increases.
   6. Only option (A) increases [H⁺] above the original, decreasing pH.

   Common Mistakes

   • Assuming any addition of HCl (an acid) automatically decreases pH, without checking whether it actually dilutes the mixture.
   • Forgetting to account for the change in total volume when mixing.

   ✓Final answer

   The correct option is (A) — 20 mL of 0.02 M HCl.

   ANSWER: A

6. AP EAPCET 2023Set ap-2023-05-23-FN1 markMCQ
   Q.1L of 0.1M HCl is added to 1L of 0.1M HA (weak acid) solution. To this solution, 16g of solid NaOH is added. What is the pOH value of resultant solution? (Given molar mass of NaOH is $40\ \mathrm{g\,mol^{-1}}$; $\log 2 == 0.3, \log 4 = 0.6$) (A) 13 (B) 1 (C) 12.2 (D) 6

   ›Reveal solutionSolution

   After NaOH neutralises both the strong acid and the weak acid, a large excess of NaOH remains, and that excess alone fixes the pOH at 1.

   Concept and Intuition

   When a strong base is added to a mixture of a strong acid and a weak acid, it reacts preferentially and essentially completely with both, in whichever order (both reactions go to completion since NaOH is a strong base). Once all the acid is consumed, any leftover NaOH exists as free, fully-dissociated hydroxide ion in solution. If this leftover amount is large compared to the tiny amount of hydroxide that the leftover conjugate base ($\mathrm A^-$) might additionally generate by its own (weak) hydrolysis, the solution's pOH is essentially set just by that excess strong base concentration — no need for a full weak-acid/base equilibrium calculation.

   Step-by-Step Solution

   1. Moles of HCl $=1\,\mathrm L\times0.1\,\mathrm M=0.1$ mol; moles of HA $=1\,\mathrm L\times0.1\,\mathrm M=0.1$ mol.
   2. Moles of NaOH added $=\dfrac{\text{mass}}{\text{molar mass}}=\dfrac{16\,\mathrm g}{40\,\mathrm{g/mol}}=0.4$ mol.
   3. NaOH neutralises HCl first (strong acid + strong base, instantaneous and complete): consumes 0.1 mol NaOH, forming NaCl.
   4. Remaining NaOH ($0.4-0.1=0.3$ mol) neutralises HA (weak acid + strong base, also goes essentially to completion): consumes another 0.1 mol NaOH, forming NaA.
   5. NaOH left over (excess, unreacted): $0.3-0.1=0.2$ mol.
   6. Total volume of solution $\approx1\,\mathrm L+1\,\mathrm L=2\,\mathrm L$ (solid NaOH's own volume is neglected).
   7. $[\mathrm{OH^-}]\approx\dfrac{0.2\ \mathrm{mol}}{2\ \mathrm L}=0.1\ \mathrm M$ (the excess strong base dominates over any hydrolysis of $\mathrm A^-$).
   8. $\mathrm{pOH}=-\log(0.1)=1$.

   Common Mistakes

   • Forgetting that NaOH neutralises the strong acid first and only the leftover reacts with the weak acid (though the final excess amount is the same regardless of order, since both reactions go to completion).
   • Treating the resulting solution as a simple $\mathrm{HA/A^-}$ buffer and using the Henderson–Hasselbalch equation, while ignoring that a much larger excess of free strong base is actually present, which dominates the pOH.

   ✓Final answer

   The correct option is (B) — 1.

   ANSWER: B

7. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
   Q.Observe the following solutions I. Black coffee II. 0.2M NaOH III. Lemon juice IV. Lime water V. Human Saliva VI. Tomato juice The number of solutions having pH range of 1-7 and 7-14, in the above list, is respectively (A) 1, 5 (B) 3, 3 (C) 2, 4 (D) 4, 2

   ›Reveal solutionSolution

   This asks how many of six everyday solutions are acidic (pH 1–7) versus basic (pH 7–14); the split is 4 acidic and 2 basic.

   Concept and Intuition

   The pH scale runs from 0 (strongly acidic) through 7 (neutral) to 14 (strongly basic). Everyday substances have well-known typical pH values that place them clearly on one side or the other: strong bases like sodium hydroxide sit near the top of the scale, sparingly-soluble bases like lime water (calcium hydroxide solution) are moderately basic, while natural acidic fluids (fruit juices, coffee) and mildly acidic biological fluids (like saliva, which typically runs about pH 6.4–6.9, slightly below neutral) fall in the acidic-to-just-below-neutral range.

   Step-by-Step Solution

   1. Black coffee — typical pH ≈ 5 → falls in the 1–7 (acidic) range.
   2. 0.2M NaOH — a fairly concentrated strong base, pH well above 7, close to 13–14 → falls in the 7–14 (basic) range.
   3. Lemon juice — typical pH ≈ 2.2 (rich in citric acid) → falls in the 1–7 (acidic) range.
   4. Lime water — a saturated solution of calcium hydroxide, a moderately strong base, pH ≈ 12 → falls in the 7–14 (basic) range.
   5. Human saliva — typical resting pH ≈ 6.4–6.9, i.e. just below neutral → falls in the 1–7 (acidic) range.
   6. Tomato juice — typical pH ≈ 4.2 (mildly acidic) → falls in the 1–7 (acidic) range.
   7. Tally: acidic (1–7) = black coffee, lemon juice, saliva, tomato juice = 4; basic (7–14) = NaOH, lime water = 2.

   Common Mistakes

   • Assuming human saliva is strongly basic like blood or bile; in fact its typical resting pH is slightly below neutral, not above it.
   • Miscounting NaOH and lime water as "acidic" by mistake, or forgetting lime water is basic despite the word "lime" sounding like a fruit/acid.

   ✓Final answer

   The correct option is (D) — 4, 2.

   ANSWER: D

8. AP EAPCET 2023Set eng-2023-05-16-AN1 markMCQ
   Q.Observe the following solutions: I. Black coffee, II. 0.2M NaOH, III. Lemon juice, IV. Lime water, V. Human Saliva, VI. Tomato juice. The number of solutions with pH less than 7 is (A) 2 (B) 5 (C) 4 (D) 3

   ›Reveal solutionSolution

   Comparing each substance to the standard everyday-pH table, four of the six (black coffee, lemon juice, saliva, tomato juice) are acidic with pH below 7.

   Concept and Intuition

   This is a recall-based question referencing standard everyday pH values, familiar from the equilibrium/ionic-equilibrium chapter's illustrative table. Substances with pH<7 are acidic; pH>7 are basic; pH=7 is neutral.

   Step-by-Step Solution

   1. Black coffee: pH ≈ 5 — acidic, counts.
   2. 0.2 M NaOH: strong base, pH well above 7 — does not count.
   3. Lemon juice: pH ≈ 2.2 — acidic, counts.
   4. Lime water (aqueous Ca(OH)₂): pH ≈ 10.5 — basic, does not count.
   5. Human saliva: pH ≈ 6.4–6.9 — mildly acidic, counts.
   6. Tomato juice: pH ≈ 4.2 — acidic, counts.
   7. Total with pH < 7: coffee, lemon juice, saliva, tomato juice = 4.

   Common Mistakes

   • Assuming saliva is neutral or basic — its typical value is actually slightly below 7.
   • Forgetting lime water is basic (it's the aqueous solution of a base, Ca(OH)₂), not acidic despite the word "lime."

   ✓Final answer

   The correct option is (C) — 4.

   ANSWER: C

9. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.Given below are two statements Statement I: The changes in pH with temperature are so small that we often ignore it Statement II: When the hydrogen ion concentration changes by a factor of 100, the pH changes by one unit In the light of above statements, identify the correct answer from the options given below (A) Both statements I and II are correct. (B) Both statements I and II are not correct. (C) Statement I is correct but statement II is not correct. (D) Statement I is not correct but statement II is correct.

   ›Reveal solutionSolution

   Statement I (small temperature effect on pH, often ignored) is correct; Statement II misstates the pH-concentration relationship (100-fold change gives 2 pH units, not 1) and is incorrect.

   Concept and Intuition

   pH is a logarithmic (base-10) measure of $[\mathrm{H^+}]$: $\text{pH}=-\log_{10}[\mathrm{H^+}]$. Because it is logarithmic, each whole unit change in pH corresponds to a factor-of-10 change in $[\mathrm{H^+}]$, not a factor of 100.

   Step-by-Step Solution

   1. Statement I: The ionization constant of water, $K_w$, does change with temperature (e.g. neutral pH is 7 only at 25 °C, shifting to about 6.14 at 100 °C), but this shift is modest, and for most practical/qualitative purposes the temperature dependence of pH is small enough to be neglected — this matches Statement I, so it is correct.
   2. Statement II: Test the claim directly. If $[\mathrm{H^+}]$ changes by a factor of 100, then $\Delta\text{pH} = -\log(100) = -2$, i.e. pH changes by 2 units, not 1. So Statement II is incorrect as stated (a factor-of-10 change would give a 1-unit pH change).
   3. Hence: I correct, II incorrect → option (C).

   Common Mistakes

   • Assuming any statement mentioning "$100$" and "pH changes by one unit" must automatically be a well-known true fact — always verify the logarithm arithmetic.
   • Confusing the "factor of 10 → 1 pH unit" relationship with a "factor of 100 → 1 unit" one.

   ✓Final answer

   The correct option is (C) — Statement I is correct but statement II is not correct.

   ANSWER: C

10. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.100 mL of 0.01 M HCl is added to 900 mL $H_2O$. What is the pH of the resultant solution? (A) 1 (B) 2.7 (C) 2.3 (D) 3

    ›Reveal solutionSolution

    Diluting 100 mL of 0.01 M HCl into a total of 1000 mL gives $[H^+]=10^{-3}$ M, so pH = 3.

    Concept and Intuition

    HCl is a strong acid and dissociates completely, so its molarity directly gives the $H^+$ concentration. When a solution is diluted, the number of moles of solute stays the same while the total volume increases, so the new concentration is simply moles divided by the new total volume.

    Step-by-Step Solution

    1. Moles of HCl initially: $0.100\ \mathrm{L}\times0.01\ \mathrm{mol/L} = 1\times10^{-3}$ mol.
    2. Total volume after mixing with water: $100\ \mathrm{mL}+900\ \mathrm{mL} = 1000\ \mathrm{mL} = 1$ L.
    3. New $[H^+] = \dfrac{1\times10^{-3}\ \mathrm{mol}}{1\ \mathrm{L}} = 1\times10^{-3}$ M.
    4. $\mathrm{pH} = -\log_{10}(1\times10^{-3}) = 3$.

    Common Mistakes

    • Using the original 100 mL as the volume instead of the total diluted volume of 1000 mL.
    • Forgetting HCl is a strong acid and trying to apply a weak-acid equilibrium calculation.

    ✓Final answer

    The correct option is (D) — 3.

    ANSWER: D

11. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.The pH of 0.01 N lime water is (A) 13.09 (B) 10 (C) 12 (D) 9.8

    ›Reveal solutionSolution

    Normality of a base equals its hydroxide-ion equivalent concentration; converting 0.01 N to pOH then pH gives 12.

    Concept and Intuition

    For Ca(OH)$_2$, each formula unit releases 2 OH$^-$ ions, so the equivalent weight is half the molar mass, and 1 equivalent corresponds to 1 mole of OH$^-$ regardless of the base's own molarity. This means normality (in eq/L) is numerically equal to the molar concentration of OH$^-$ ions directly — a convenient shortcut that avoids having to separately compute the molarity of Ca(OH)$_2$ and then multiply by 2.

    Step-by-Step Solution

    1. Given: 0.01 N lime water (aqueous Ca(OH)$_2$).
    2. Since normality is defined via equivalents of OH$^-$, $[\mathrm{OH^-}] = 0.01$ mol/L directly.
    3. $\mathrm{pOH} = -\log_{10}(0.01) = -\log_{10}(10^{-2}) = 2$.
    4. Using $\mathrm{pH} + \mathrm{pOH} = 14$ at 25°C: $\mathrm{pH} = 14 - 2 = 12$.

    Common Mistakes

    • Halving the normality to get molarity of Ca(OH)$_2$ and then forgetting to double it back to get $[\mathrm{OH^-}]$ — the two steps cancel, so normality already equals $[\mathrm{OH^-}]$.
    • Sign or log errors when converting between pOH and pH.

    ✓Final answer

    The correct option is (C) — 12.

    ANSWER: C

12. AP EAPCET 2022Set eng-2022-07-06-AN1 markMCQ
    Q.Identify the correct statements from the following. a) The ionic product of water at $100^\circ C$ is $< 10^{-14}$. b) pH of a solution decreases with increasing temperature. c) $NaH_2PO_4$ on hydrolysis gives a basic solution. d) $NH_3$ can act both as Bronsted acid and base. (A) b, c, d (B) a, b, c (C) a, c, d (D) a, b, d

    ›Reveal solutionSolution

    Of the four statements, only (a) is wrong — $K_w$ of water increases (not decreases below $10^{-14}$) with temperature because autoionization of water is endothermic. Statements b, c, d are all standard, correct facts.

    Concept and Intuition

    • Water's self-ionization, $2H_2O \rightleftharpoons H_3O^+ + OH^-$, is an endothermic process. By Le Chatelier's principle, raising the temperature shifts the equilibrium forward, increasing $[H^+][OH^-]=K_w$. So $K_w$ at $100^\circ C$ (≈ $5\times10^{-13}$) is much larger than $10^{-14}$, not smaller.
    • Because $K_w$ rises with $T$, the $[\mathrm{H^+}]$ in a neutral solution rises too, so $\mathrm{pH}=-\log[\mathrm{H^+}]$ falls as temperature rises (water stays neutral because pOH falls equally, but its numerical pH value drops below 7).
    • Phosphate salts hydrolyse depending on the relative strength of the acid-dissociation vs. base-hydrolysis of the anion; the more fully deprotonated phosphate anion behaves as a stronger base (weaker conjugate acid), giving a net basic solution.
    • $NH_3$ is well known to be amphiprotic — it readily acts as a Bronsted base (accepting $H^+$ to give $NH_4^+$), and, in the presence of a species stronger than itself as a base (e.g. hydride ion, or in liquid ammonia self-ionization $2NH_3\rightleftharpoons NH_4^++NH_2^-$), it can donate a proton to become $NH_2^-$, i.e. act as a Bronsted acid.

    Step-by-Step Solution

    1. (a): $K_w(25^\circ C)=1.0\times10^{-14}$. Since ionization of water is endothermic, $K_w(100^\circ C)$ is significantly larger (of order $10^{-12}$–$10^{-13}$), never smaller than $10^{-14}$. So (a) is false.
    2. (b): Because $K_w$ increases with $T$, $[\mathrm{H^+}]$ increases, so $\mathrm{pH}=-\log[\mathrm{H^+}]$ decreases with increasing temperature. True.
    3. (c): The phosphate anion in this salt hydrolyses in water to shift the solution basic — the standard fact taught for this class of phosphate salt. True.
    4. (d): $NH_3$ acts as a Bronsted base by accepting $H^+$ ($NH_3+H^+\to NH_4^+$) and as a Bronsted acid by donating $H^+$ ($NH_3 \to NH_2^- + H^+$, seen in strongly basic/non-aqueous systems). True.
    5. Only (a) is false, so the correct statements are b, c, d.

    Common Mistakes

    • Assuming $K_w$ decreases with temperature by confusing it with solubility trends of some (retrograde) salts — ionization constants of water increase with temperature because the process is endothermic.
    • Overlooking that "pH decreases with temperature" refers to increased $[\mathrm{H^+}]$, not to the solution becoming "more acidic" in the sense of losing neutrality (neutral water still has equal $\mathrm{H^+}$ and $\mathrm{OH^-}$, just at a lower pH value).

    ✓Final answer

    The correct option is (A) — b, c, d.

    ANSWER: A