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

2024 · 30 Jan · Shift 2 · Q26
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. /2024 · 30 Jan · Shift 2 · Q26

Ionic Equilibrium question

2024 · 30 Jan · Shift 2 · Q26

JEE MainChemistryIonic EquilibriumNumerical+4 / −1
The pH\mathrm{pH}pH of an aqueous solution containing 1M1 \mathrm{M}1M benzoic acid (pKa=4.20)\left(\mathrm{pK}_{\mathrm{a}}=4.20\right)(pKa​=4.20) and 1M1 \mathrm{M}1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL300 \mathrm{~mL}300 mL of this buffer solution is ‾\underline{\hspace{2cm}}​mL\mathrm{mL}mL. (given : log⁡2=0.3\log 2=0.3log2=0.3)
Numerical answer
View written solutionFree

Correct answer: 100

  1. Use the Henderson–Hasselbalch equation for the buffer:
pH=pKa+log⁡[salt][acid]\mathrm{pH} = \mathrm{p}K_a + \log \frac{[\text{salt}]}{[\text{acid}]}pH=pKa​+log[acid][salt]​

Given:

pH=4.5,pKa=4.20\mathrm{pH}=4.5, \qquad \mathrm{p}K_a=4.20pH=4.5,pKa​=4.20

So,

4.5=4.2+log⁡[sodium benzoate][benzoic acid]4.5 = 4.2 + \log \frac{[\text{sodium benzoate}]}{[\text{benzoic acid}]}4.5=4.2+log[benzoic acid][sodium benzoate]​ 0.3=log⁡[salt][acid]0.3 = \log \frac{[\text{salt}]}{[\text{acid}]}0.3=log[acid][salt]​

Given log⁡2=0.3\log 2 = 0.3log2=0.3, therefore

[salt][acid]=2\frac{[\text{salt}]}{[\text{acid}]} = 2[acid][salt]​=2
  1. Let the volume of 1 M1\,\text{M}1M benzoic acid used be VaV_aVa​ mL. Let the volume of 1 M1\,\text{M}1M sodium benzoate used be VsV_sVs​ mL.

Total buffer volume is 300300300 mL, so

Va+Vs=300V_a + V_s = 300Va​+Vs​=300

Since both solutions are 1 M1\,\text{M}1M, the ratio of concentrations in the final mixture equals the ratio of moles, which equals the ratio of volumes:

[salt][acid]=VsVa=2\frac{[\text{salt}]}{[\text{acid}]} = \frac{V_s}{V_a} = 2[acid][salt]​=Va​Vs​​=2

Thus,

Vs=2VaV_s = 2V_aVs​=2Va​
  1. Substitute into the total volume equation:
Va+2Va=300V_a + 2V_a = 300Va​+2Va​=300 3Va=3003V_a = 3003Va​=300 Va=100 mLV_a = 100\,\text{mL}Va​=100mL
  1. Therefore, the volume of benzoic acid solution is
100 mL\boxed{100\,\text{mL}}100mL​
PreviousNext

More from Ionic Equilibrium

  • The titration curve of weak acid vs. strong base with phenolphthalein as indictor) is shown below. The Kphenolphthalein ​=4×10−10. Given: log2=0.3 The number of following statement/s which is/are… Includes diagram2023 · Numerical
  • The solubility product of BaSO4​ is 1×10−10 at 298 K. The solubility of BaSO4​ in 0.1 M K2​SO4​(aq) solution is ​×10−9 g L−1…2023 · Numerical
  • 25 mL of silver nitrate solution (1M) is added dropwise to 25 mL of potassium iodide (1.05M) solution. The ion(s) present in very small quantity in the solution is/are :2023 · MCQ
  • An analyst wants to convert 1 L HCl of pH=1 to a solution of HCl of pH 2. The volume of water needed to do this dilution is ​mL. (Nearest integer)2023 · Numerical
  • 25.0 mL of 0.050 M Ba(NO3​)2​ is mixed with 25.0 mL of 0.020 M NaF.KSp​ of BaF2​ is 0.5×10−6 at 298 K…2023 · Numerical
  • 20 mL of 0.1 M NaOH is added to 50 mL of 0.1 M acetic acid solution. The pH of the resulting solution is ​×10−2(Nearest integer) Given : pKa(CH3​COOH)=4.76…2023 · Numerical
  • Which of the following statement(s) is/are correct? (A) The pH of 1×10−8 M HCl solution is 8 . (B) The conjugate base of H2​PO4−​ is HPO42−​. (C) Kw​…2023 · MCQ
  • The dissociation constant of acetic acid is x×10−5. When 25 mL of 0.2 M CH3​COONa solution is mixed with 25 mL of 0.02 M CH3​COOH solution, the pH of the resultant solution is found to be equal to 5. The…2023 · Numerical