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Equilibrium MCQs for NEET — Chemistry Questions with Answers

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Equal volumes of three basic solutions of 11,12 and13 are mixed in a vessel. what will be the $ H^+ $ ion concentration in the mixture ?

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Explanation

$ [H ^ + ] ion concentrations are 10 ^{-11}, 10 ^{-12} and 10 ^{-13} on mixing equal volumes, [H^+] $ in final solution $ = { 10 ^ {-11} \times 10 ^ {-12} \times 10 ^ {-13} \over 3 } = { 10 \times 10 ^ {-12} + 1 \times 10 ^ {-12} + 0.1 \times 10 ^ {-12} \over 3 } $ $ = 11.1 \times 10 ^ {-12 } \over 3 } = 3.7 \times 10 ^ {-12} M $

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Column I (buffer solution ) (A) $ 0.1 M CH_3 COOH + 0.01 M CH_3 COONa $ (B) $ 0.01 M CH_3 COOH + 0.1 M CH_3 COO Na $ ('C) $ 0.1 M CH_3 COOH + 0.1 M CH_3 COO Na $ (D) $ 0.1 M CH_3 COONH_4 $ $ ( P ^ {ka} of CH_3 COOH + p ^ {kb} of NH_4 OH = 4.8 ) $ Column II (p) 3.8 (q) 5.8 (v) 7.0 (s) 4.8

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Explanation

(A) $ P^H = pka + log {[Salt] \over [acid] } = 4.8 + log { 0.1 \over 0.01 } = 4.8 + 1 = 5.8 $ (B) $ P^H = 4.8 + log {0.01 \over 0.1 } = 4.8 + log 10^{-1} = 4.8 -1 = 3.8 $ (C)$ P^H = 4.8 + log {0.1 \over 0.1 } = 4.8 + 0 = 4.8 $ (D) $ P^H = 7 + 1/2 (pka - pkb) = 7 + 1/1 (4.8 - 4.8) =7.0 $

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$ P^H $ of a soda water bottle is

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Explanation

Soda water contains weak acid $ H_2CO_3 So its P^H \lt 7 $

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Statement :1 $P^H of 10^{-8} HClsolution is not equal to 8 Statement :2 HCI does not dissociate properly in verydilute solution.

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Explanation

Statement 1 is true: The pH of a 10^{-8} M HCl solution is not equal to 8 because in such a dilute solution, the autoionization of water also contributes significantly to the hydrogen ion concentration. Statement 2 is false: HCl is a strong acid and it dissociates completely in water, even in very dilute solutions.

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For preparing a buffer solution of $ P^H 6 by mixing sodium acetate and acetic acid , the ratio of concentration of salt and acid should be, (Ka=10^{-5} $ )

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Explanation

$ P^H = p^K a + log {[Salt] \over [acid] } $ $ 6 = 5 + log { [Salt] \over [acid] } $ $ \therefore log {[Salt] \over [acid]} = 1 or { [salt] \over [acid]} = 10 $

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The $ K_{SP} of CuS, Ag_2 S and HgSare 10 ^ {-32}, 4 \times 10 ^ {-45} and 10^{-54} $ respectively. The solubility of these sulphides are in order of

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Explanation

$ Solubility of Cus = (ksp) ^ {1/2} = ( 1 \times 10 ^ {-32} ) ^ {1/2} = 1 \times 10 ^ {-16} M $ $ Solubility of Ag_2 S = ( { ksp \over 4} ) ^ { 1/3} = ( { 4 \times 10 ^ {-45 } \over 4 } ) ^ {1/3} = 1\times 10 ^ {-15} M $ $ Solubility of HgS = (Ksp) ^ {1/2} = ( 4 \times 10 ^ {-54 } ) ^ {1/5} = 1 \times 10 ^ {-27 } M $

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The $ P^Hof neutral water is 6.8 .Then the temperature of H_2O $

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Explanation

$ At 25 ^\circ C temp P^H of H_2 O = 7 [H^+] = 10 ^{-7} M $ $ P^H = 6.8 means P^H \lt 7 \therefore [ H^+ ] = 10 ^ {-7} M $ $ Self ionisation of H_2O is endothermic so by increasing temp [H^+] $ ion increases.

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On adding 0.1 M solution each of $ Ag ^+,Ba ^{2+} and Ca ^ {2+} ions. in a Na_2SO_4 solution. species first precipitated is (K_{SP} of BaSO_4 =10 ^{-11} , K_{SP} of CaSO_4 =10 ^{- 6} , K of Ag_2SO_4 = 10 ^{-5 } $

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Explanation

$ Ksp for BaSO_4 = [ Ba^{2+} ] [ SO_4 ^ {2-} ] $ $ 10 ^ {-11} = 0.1 \times [SO_4 ^ {2-} ] $ $ \therefore [ SO_4 ^ {2-} ] = 10 ^ {-10 } M $ $ ksp for CaSO_4 = [ Ca^{2+} ] [SO_4 ^ {2-} ] $
$ { 10 ^{-6} \over 0.1 } = 10 ^ {-5} M = [ SO_4 ^ {2-} ] $ $ ksp for Ag_2 SO_4 = [ Ag^+] ^ 2 [SO_4 ^ {2-} ] $ $ { 10 ^ {-5} \over (0.1) ^ 2 } = 10 ^ {-3} M = [SO_4 ^ {2-} ] $ $ As [SO_4 ^ {2-} ] = 10 ^ {-10} M in BaSO_4 (least value ) $ it can be precipitated first

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The solubility of $ A_2 B_3 is x mol L^{-1} $ . It solubility product is

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Explanation

$ A_2 B_{3(s)} \rightleftharpoons 2A^ {3+} + 3 B ^ {2-} $ 2x 3x $ ksp = [A^{3+} ] ^2 [ B^{2-} ] ^ 3 $ $ = (2x) ^ 2 \times (3x)^3 $ $ = 4 x ^ 2 \times 27 x^3 = 108 x ^ 5 $

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How much volume of 0.1 M $CH_3COOH should be added to 50ml of 0.2 M CH_3COONa If we want to prepare a buffer Solution of P^H $ 4.91. given pKa for acetic acid is 4.76

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Explanation

$ P ^ H = P ^ K a + log { [salt ] \over [acid] } $ $ 4.91 = 4.76 + log { [ salt ] \over [acid ] } $ $ \therefore log { [salt ] \over [acid ] = 0.15 $ $ \therefore { [salt ] \over [acid] } = antilog of 0.15 = 1.41 $ $ \therefore { {0.2 \over 1000} \times 50 \over {0.1 \over 1000} \times V } = 1.41 $ $ \therefore V = 70.92 ml $

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