Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Ionic Equilibrium question

2025 · 4 Apr · Shift 1 · Q21
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Chemistry
  4. /Ionic Equilibrium
  5. /2025 · 4 Apr · Shift 1 · Q21

Ionic Equilibrium question

2025 · 4 Apr · Shift 1 · Q21

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The pH of a 0.01 M weak acid HX(Ka=4×10−10)\mathrm{HX}\left(\mathrm{K}_a=4 \times 10^{-10}\right)HX(Ka​=4×10−10) is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as x×10−4Mx \times 10^{-4} \mathrm{M}x×10−4M. The value of xxx is ‾\underline{\hspace{2cm}}​ (nearest integer)
Numerical answer
View written solutionFree

Correct answer: 25

  1. Initial solution check

Given weak acid HX\mathrm{HX}HX with: Ka=4×10−10K_a = 4\times 10^{-10}Ka​=4×10−10 Initial concentration: C1=0.01 MC_1 = 0.01\,\text{M}C1​=0.01M Observed pH =5=5=5, so [H+]=10−5 M[\mathrm{H}^+] = 10^{-5}\,\text{M}[H+]=10−5M

For a weak acid, Ka=[H+][X−][HX]K_a = \frac{[\mathrm{H}^+][\mathrm{X}^-]}{[\mathrm{HX}]}Ka​=[HX][H+][X−]​ Since dissociation is small, [X−]≈[H+]=10−5[\mathrm{X}^-] \approx [\mathrm{H}^+] = 10^{-5}[X−]≈[H+]=10−5 [HX]≈0.01[\mathrm{HX}] \approx 0.01[HX]≈0.01 Thus, Ka≈(10−5)(10−5)10−2=10−8K_a \approx \frac{(10^{-5})(10^{-5})}{10^{-2}} = 10^{-8}Ka​≈10−2(10−5)(10−5)​=10−8 which is much larger than the given 4×10−104\times 10^{-10}4×10−10. So the given pH =5=5=5 for the original solution is not consistent with ordinary weak-acid dissociation alone.

This suggests we should use the dilution condition asked in the problem with the given KaK_aKa​ and final pH.

  1. After dilution

Final pH =6=6=6, hence [H+]=10−6 M[\mathrm{H}^+] = 10^{-6}\,\text{M}[H+]=10−6M Let the new concentration of acid be CCC.

For weak acid HX\mathrm{HX}HX: Ka=[H+][X−][HX]K_a = \frac{[\mathrm{H}^+][\mathrm{X}^-]}{[\mathrm{HX}]}Ka​=[HX][H+][X−]​ At equilibrium, [H+]=10−6,[X−]≈10−6,[HX]≈C−10−6[\mathrm{H}^+] = 10^{-6}, \quad [\mathrm{X}^-] \approx 10^{-6}, \quad [\mathrm{HX}] \approx C-10^{-6}[H+]=10−6,[X−]≈10−6,[HX]≈C−10−6 Therefore, 4×10−10=(10−6)(10−6)C−10−64\times 10^{-10} = \frac{(10^{-6})(10^{-6})}{C-10^{-6}}4×10−10=C−10−6(10−6)(10−6)​

So, C−10−6=10−124×10−10C-10^{-6} = \frac{10^{-12}}{4\times 10^{-10}}C−10−6=4×10−1010−12​ C−10−6=14×10−2=2.5×10−3C-10^{-6} = \frac{1}{4}\times 10^{-2} = 2.5\times 10^{-3}C−10−6=41​×10−2=2.5×10−3 Hence, C=2.5×10−3+10−6=2.501×10−3 MC = 2.5\times 10^{-3} + 10^{-6} = 2.501\times 10^{-3}\,\text{M}C=2.5×10−3+10−6=2.501×10−3M

Now write this as C=x×10−4C = x\times 10^{-4}C=x×10−4 Then, x=2.501×10−310−4=25.01x = \frac{2.501\times 10^{-3}}{10^{-4}} = 25.01x=10−42.501×10−3​=25.01 Nearest integer: x=25x = 25x=25

  1. Final answer

25\boxed{25}25​

  1. Comparison with stored answer

Stored correct answer = 252525

My derived answer matches it.

PreviousNext

More from Ionic Equilibrium

  • The molar conductance of an infinitely dilute solution of ammonium chloride was found to be 185 S cm2 mol−1 and the ionic conductance of hydroxyl and chloride ions are 170 and 70 S cm2 mol−1…2025 · Numerical
  • xmg of Mg(OH)2​( molar mass =58) is required to be dissolved in 1.0 L of water to produce a pH of 10.0 at 298 K . The value of x is ​ mg. (Nearest integer) (Given : Mg(OH)2​…2025 · Numerical
  • An aqueous solution of HCl with pH 1.0 is diluted by adding equal volume of water (ignoring dissociation of water). The pH of HCl solution would ( Given log2=0.30)2025 · MCQ
  • The percentage dissociation of a salt (MX3​) solution at given temperature (van't Hoff factor i=2 ) is ​ %(Nearest integer)2025 · Numerical
  • One litre buffer solution was prepared by adding 0.10 mol each of NH3​ and NH4​Cl in deionised water. The change in pH on addition of 0.05 mol of HCl to the above solution is ​×10−2…2025 · Numerical
  • The molar solubility(s) of zirconium phosphate with molecular formula (Zr4+)3​(PO43−​)4​ is given by relation :2025 · MCQ
  • Which of the following happens when NH4​OH is added gradually to the solution containing 1 M A2+ and 1MB3+ ions? Given : Ksp ​[A(OH)2​]=9×10−10…2025 · MCQ
  • If 1 mM solution of ethylamine produces pH=9, then the ionization constant (Kb​) of ethylamine is 10−x. The value of x is ​ (nearest integer). [The degree of…2025 · Numerical