Ionic Equillibrium — NEET UG practice

64 questions

Practice NEET UG Ionic Equillibrium questions free — each with a detailed solution, graded instantly. Nothing is saved; log in to track your accuracy and build a streak.

Start free practice

Sample questions with solutions

Q1 · 2026

At 298 K , a certain buffer solution contains equal concentrations of X−X^{-}and HXHX, KbK_{b} for X−X^{-} is 10−1010^{-10}. What is the pH of this buffer solution?

  • A.

    2

  • B.

    4

  • C.

    6

  • D.

    10

Answer: B
  1. Since X−X^{-} is the conjugate base of the weak acid HXHX, the two constants are related by Ka×Kb=KwK_a \times K_b = K_w.
Ka=KwKb=10−1410−10=10−4K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4}
  1. For a buffer made of a weak acid and its conjugate base, use the Henderson–Hasselbalch equation.
pH=pKa+log⁡[X−][HX]pH = pK_a + \log\frac{[X^{-}]}{[HX]}
  1. Since the buffer contains equal concentrations of X−X^{-} and HXHX, the ratio inside the log is 11, and log⁡1=0\log 1 = 0.
pH=pKa+0pH = pK_a + 0

Given Ka=10−4K_a=10^{-4}, so pKa=4pK_a=4.

pH=4pH = 4

Hence, the answer is Option B.

Q2 · 2026

Phenolphthalein is used as an indicator for the titration of sodium hydroxide solution against a standard solution of oxalic acid. The colour change that is observed at an alkaline pH close to the equivalence point during this titration is:

  • A.

    pinkish red to yellow

  • B.

    yellow to pinkish red

  • C.

    pink to colourless

  • D.

    colourless to pink

Answer: D

  1. Recall the colour behaviour of phenolphthalein: it stays colourless in acidic and neutral solutions, and turns pink once the solution becomes sufficiently basic.

  2. Since oxalic acid (a weak acid) is titrated against sodium hydroxide (a strong base), the salt formed at the equivalence point undergoes hydrolysis and makes the solution slightly alkaline, matching phenolphthalein's colour-change range (pH ≈8.2\approx 8.2–10.010.0).

  3. So during the titration, as excess NaOHNaOH is added past the equivalence point, the solution turns basic and the indicator responds.

Hence, the colour change observed is colourless to pink, so the answer is Option D.

Q3 · 2026

The correct order of solubility of the given salts in water at 298 K298\ \mathrm{K} is

SaltKspK_{sp} at 298 K298\ K
AgBrAgBr5.0×10−135.0 \times 10^{-13}
Zn(OH)2Zn(OH)_21.0×10−151.0 \times 10^{-15}
Hg2Cl2Hg_2Cl_21.3×10−181.3 \times 10^{-18}
  • A.

    Zn(OH)2>AgBr>Hg2Cl2Zn(OH)_2>AgBr>Hg_2Cl_2

  • B.

    Hg2Cl2>Zn(OH)2>AgBrHg_2Cl_2>Zn(OH)_2>AgBr

  • C.

    AgBr>Zn(OH)2>Hg2Cl2AgBr>Zn(OH)_2>Hg_2Cl_2

  • D.

    Hg2Cl2>AgBr>Zn(OH)2Hg_2Cl_2>AgBr>Zn(OH)_2

Answer: A
  1. For AgBrAgBr, since it dissociates into one Ag+Ag^{+} and one Br−Br^{-}, Ksp=s2K_{sp}=s^2, where ss is the molar solubility.

Given Ksp=5.0×10−13K_{sp}=5.0\times10^{-13}, so

s2=5.0×10−13s^2 = 5.0\times10^{-13}

Hence

s=5.0×10−13=7.07×10−7 mol/Ls = \sqrt{5.0\times10^{-13}} = 7.07\times10^{-7}\ \mathrm{mol/L}
  1. For Zn(OH)2Zn(OH)_2, it dissociates into one Zn2+Zn^{2+} and two OH−OH^{-}, so if solubility is ss, then [OH−]=2s[OH^-]=2s, giving Ksp=4s3K_{sp}=4s^3.

Given Ksp=1.0×10−15K_{sp}=1.0\times10^{-15}

4s3=1.0×10−154s^3 = 1.0\times10^{-15}

Therefore

s=6.3×10−6 mol/Ls = 6.3\times10^{-6}\ \mathrm{mol/L}
  1. For Hg2Cl2Hg_2Cl_2, it dissociates into one Hg22+Hg_2^{2+} and two Cl−Cl^{-}, so again Ksp=4s3K_{sp}=4s^3.

Given Ksp=1.3×10−18K_{sp}=1.3\times10^{-18}

4s3=1.3×10−184s^3 = 1.3\times10^{-18}

So

s=6.9×10−7 mol/Ls = 6.9\times10^{-7}\ \mathrm{mol/L}
  1. Comparing the three solubility values: 6.3×10−6>7.07×10−7>6.9×10−76.3\times10^{-6} > 7.07\times10^{-7} > 6.9\times10^{-7}.

Hence, the order is Zn(OH)2>AgBr>Hg2Cl2Zn(OH)_2>AgBr>Hg_2Cl_2, so the answer is Option A.

Do more with a free account

  • Take it as a timed mock test
  • Build a daily streak
  • Track accuracy & progress
  • Save questions to your library
Sign up freeLog in