The pH of a 0.01mathrm~M weak acid HX (K_a = 4 times 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 times 10^-4mathrm~M. The value of x is _______ (nearest integer).

Numerical Answer Type:
Enter a numerical value Answer: 25 to 25 +4 marks

Solution & Explanation

### Related Formula HX_(aq) rightleftharpoons H^+_(aq) + X^-_(aq) K_a = frac[H^+][X^-][HX] = frac(Calpha)^2C(1-alpha) approx Calpha^2 ### Core Logic Official Answer Path Analysis: When the solution is diluted until mathrmpH = 6, the hydronium ion concentration becomes: [H^+] = 10^-6mathrm~M = C_textnewalpha_textnew Applying the equilibrium constant expression without approximations for very high dilutions: K_a = fracCalpha^21-alpha = frac[H^+]alpha1-alpha = 4 times 10^-10 frac10^-6 cdot alpha1-alpha = 4 times 10^-10 implies 10^4 alpha = 4(1-alpha) 2500alpha = 1 - alpha implies 2501alpha = 1 implies alpha approx frac12500 Now, substitute alpha back to isolate the absolute concentration parameter C_textnew: C_textnewalpha = 10^-6 implies C_textnew cdot left(frac12500right) = 10^-6 C_textnew = 2500 times 10^-6 = 25 times 10^-4mathrm~M Therefore, comparing with x times 10^-4mathrm~M yields x = 25. ### Pattern Recognition When dealing with extreme dilution states where alpha becomes large, you must avoid the standard (1-alpha) approx 1 simplification step to ensure mathematically accurate answers. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Ionic Equilibrium

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Ionic Equilibrium

More Ionic Equilibrium Previous-Year Questions — Page 5

Q76 jee_main_2024_27_jan_morning Salt Hydrolysis
Given below are two statements: Statement (I): Aqueous solution of ammonium carbonate is basic. Statement (II): Acidic/basic nature of salt solution of a salt of weak acid and weak base depends on K_a and K_b value of acid and the base forming it. In the light of the above statements, choose the most appropriate answer from the options given below :
  • A. Both Statement I and Statement II are correct
  • B. Statement I is correct but Statement II is incorrect
  • C. Both Statement I and Statement II are incorrect
  • D. Statement I is incorrect but Statement II is correct

Solution

### Related Formula pH of a weak acid-weak base salt system: textpH = 7 + frac12(textpK_a - textpK_b) ### Core Logic Ammonium carbonate, (textNH_4)_2textCO_3, is formed from a weak acid (textH_2textCO_3, K_a approx 4.3 times 10^-7) and weak base (textNH_4textOH, K_b approx 1.8 times 10^-5). Since K_b > K_a, the aqueous medium accumulates an excess of hydroxyl particles over hydronium, forming a basic system (textpH > 7). Both statements are structurally accurate descriptions. ### Chapter Mix Class 11 Chemistry: Equilibrium
Q84 jee_main_2024_29_jan_morning Kp and Kc Relationship
For the reaction mathrmN_2mathrmO_4(mathrmg) rightleftharpoons 2mathrmNO_2(mathrmg) mathrmK_p = 0.492 atm at 300mathrmK . mathrmK_c for the reaction at same temperature is \_\_\_\_\_\_ times 10^-2 . (textGiven: R = 0.082 text L atm mathrmmol^-1 textK^-1)
Numerical Answer. Answer: 2 to 2

Solution

### Related Formula K_p = K_c cdot (RT)^Delta n_g ### Core Logic For the given gaseous equilibrium reaction: N_2O_4(g) rightleftharpoons 2NO_2(g) First, find the change in the number of moles of gas (Delta n_g): Delta n_g = n_p - n_r = 2 - 1 = 1 ### Step 1: Calculation Substitute the given values into the K_p - K_c relationship: K_p = 0.492 R = 0.082 T = 300text K 0.492 = K_c cdot (0.082 times 300)^1 K_c = frac0.4920.082 times 300 K_c = frac0.49224.6 K_c = 0.02 Converting to the requested format (x times 10^-2): K_c = 2 times 10^-2 So, the value is 2. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q85 jee_main_2024_30_january_evening Buffer Solutions
The pH of an aqueous solution containing 1M benzoic acid (pK_a = 4.20) and 1M sodium benzoate is 4.5. The volume of benzoic acid solution in 300 mL of this buffer solution is ______ mL.
Numerical Answer. Answer: 100 to 100

Solution

### Related Formula Henderson-Hasselbalch Equation for Acidic Buffers: mathrmpH = pK_a + log left( frac[textSalt][textAcid] right) ### Core Logic Let the volume of 1M Benzoic acid be V_a mL and the volume of 1M Sodium benzoate be V_s mL. Total volume = V_s + V_a = 300\,textmL. Millimoles of acid = 1 times V_a = V_a Millimoles of salt = 1 times V_s = V_s Applying Henderson's Equation: 4.5 = 4.2 + log left(fracV_sV_aright) ### Step 1: Calculate Volume Ratio log left(fracV_sV_aright) = 4.5 - 4.2 = 0.3 Since log 2 approx 0.3, we have: fracV_sV_a = 2 V_s = 2 V_a ### Step 2: Substitute and Solve We know V_s + V_a = 300 Substituting V_s = 2 V_a: 2 V_a + V_a = 300 3 V_a = 300 V_a = 100 \, textmL ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q82 jee_main_2024_30_jan_morning Solubility Product
The pH at which Mg(OH)_2 [K_sp=1times 10^-11] begins to precipitate from a solution containing 0.10text M Mg^2+ ions is
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula K_sp = [Mg^2+][OH^-]^2 pOH = -log[OH^-] pH + pOH = 14 ### Core Logic Precipitation begins just when the ionic product equals the solubility product (Q_sp = K_sp). ### Step 1: Calculating required [OH-] [Mg^2+][OH^-]^2 = 10^-11 Given [Mg^2+] = 0.10 text M 0.10 times [OH^-]^2 = 10^-11 [OH^-]^2 = 10^-10 [OH^-] = 10^-5 text M ### Step 2: Finding pH pOH = -log(10^-5) = 5 pH = 14 - pOH pH = 14 - 5 = 9 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium
Q71 jee_main_2024_31_jan_evening Equilibrium Constants (Kp and Kc)
A_(g) rightleftharpoons B_(g) + fracC2_(g) The correct relationship between K_P, alpha and equilibrium pressure P is
  • A. K_P = fracalpha^frac12P^frac12(2 + alpha)^frac12
  • B. K_P = fracalpha^frac32P^frac12(2 + alpha)^frac12(1 - alpha)
  • C. K_P = fracalpha^frac12P^frac32(2 + alpha)^frac32
  • D. K_P = fracalpha^frac12P^frac12(2 + alpha)^frac32

Solution

### Related Formula K_P = fracP_B cdot (P_C)^frac12P_A where P_i is the partial pressure of component i. ### Step 1: Setting up the ICE Table For the reaction A_(g) rightleftharpoons B_(g) + frac12 C_(g) Let initial moles of A = 1. At equilibrium: Moles of A = 1 - alpha Moles of B = alpha Moles of C = fracalpha2 Total moles at equilibrium = (1 - alpha) + alpha + fracalpha2 = 1 + fracalpha2 = frac2 + alpha2 ### Step 2: Calculating Partial Pressures Using mole fraction times Total Pressure (P): P_A = frac1 - alpha1 + fracalpha2 cdot P P_B = fracalpha1 + fracalpha2 cdot P P_C = fracfracalpha21 + fracalpha2 cdot P ### Step 3: Calculating Kp K_P = fracP_B cdot (P_C)^frac12P_A K_P = fracleft( fracalpha1 + alpha/2 P right) cdot left( fracalpha/21 + alpha/2 P right)^1/2frac1 - alpha1 + alpha/2 P K_P = fracalpha cdot (alpha/2)^1/2 cdot P^3/2(1 + alpha/2)^3/2 cdot frac1 + alpha/2(1 - alpha) P K_P = fracalpha^3/2 cdot P^1/2sqrt2 cdot (1 + alpha/2)^1/2 cdot (1 - alpha) Since 1 + alpha/2 = frac2+alpha2, the sqrt2 in denominator cancels out perfectly leaving: K_P = fracalpha^frac32 P^frac12(2 + alpha)^frac12(1 - alpha) ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Equilibrium

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