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Which of the following weakest ?

A. $ C_6 H_5 NH_2 : K_b = 3.8 \times 10 ^ {-10} $ B. $ NH_4 OH : K_b = 1.6 \times 10 ^ {-5} $ C. $ C_2 H_5 NH_2 : K_b = 5.6 \times 10 ^ {-4} $ D. $ C_9 H_7 N : K_b = 6.3 \times 10 ^ {-10} $

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Explanation

Smaller the Kb value, weaker is the base.

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On adding ammonia to water

A. Ionic product will increase B. Ionic product will decrease C. $ [ H_3 O ] ^ + $ will increase D. $ [ H_3 O ^ + ] $ will decrease

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Explanation

On adding$ NH_3 to water, [OH ^-] will increase, Kw = [H_3O^+][OH ^ - ] is constant.$ Therefore $ [H_3O ^ + ] $ will decrease

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The $ P^H$ of a buffer containing equalmolar concentration of a weak base and its chlorides is $ (K for weak base = 2 \times 10 ^ {-5} )(log 2 = 0.3010)$

A. 5 B. 9 C. 4.7 D. 9.3

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Explanation

$ \therefore P^{OH} = P^Kb + log {[Salt] \over [Base] } [Salt] = [ Base ] $ $ = - log(2 \times 10 ^ {-5} ) + log1 $ $ = - 0.3010 + 5.000 + 0 = 4.6990 $ $ P^H = 14 - p ^ {OH} = 14 - 4.6990 = 9.3010 $

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Solution of 0.1 N $NH_4OH and 0.1N NH_4Cl has P^H 9.25.Then P_{K_b}$ is

A. 9.25 B. 4.75 C. 3.75 D. 8.25

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Explanation

$ P^{OH} = p^Kb + log { [Salt] \over [Base]} $ $ P^{OH} =14 - P^H = 4.75 $
$ 4.75 = p^{kb} + log1 $ $ pk_b = 4.75 ( log 1 = 0 )

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The solubility of product barium sulphate is $ 1.5 \times 10 ^ {-9}$ at $18 ^\circ$ C Its solubility in water at $18 ^\circ C $ is

$ 1.5 \times 10 ^ {-9} mol L ^ {-1} $

$ 1.5 \times 10 ^ {-5} mol L ^ {-1} $

$ 3.9 \times 10 ^ {-9} mol L ^ {-1} $

$ 3.9 \times 10 ^ {-5} mol L ^ { -1} $

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Explanation

$ BaSO_{4(s)} \rightleftharpoons Ba^{2+} _{(aq)} + So ^ {2-}_{4(aq)} $ xmol/lit x mol/lit $ K_{SP} = [Ba^{2+} ][ So ^{2-} _ { 4} ] $ $ K_{SP} = X^2 $ $ \therefore x = (ksp) ^ { 1 /2 } = (1.5 \times 10 ^ {-9} ) ^ {1/2} = 3.9 \times 10 ^ { -5} mol L ^ { -1} $

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The least soluble compound (salt) of the following is

A. ZnS ( $ Ksp = 1.2 \times 10 ^ {-23} $ ) B. $ OH ^ - \gt Cl ^ - \gt CH_3 COO ^ - $ C. $ CsCl ($ Ksp = 1 \times 10 ^ {-12} $ D. $ PbCl_2 ( $ Ksp = 1.7 \times 10 ^ {-5} $ )

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Explanation

Hint : As value of $ K_{SP} $ is less , solubility is also less.

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What is the value of $ P^H of 0.01 M glycine solution ? For glycine Ka_1 = 4.5 \times 10 ^ {-3} and Ka_2 = 1.7 \times 10 ^ {-10} $ at 298k.

A. 3 B. 10 C. 6.1 D. 7.06

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Explanation

$ glycine (NH_2CH_2COOH)$ is more acidic than basic. overall ionisation constant $ K = K_{a1} \times K_{a2} = 4.5 \times 10 ^ {-3} \times 1.7 \times 10 ^ {-10} $ $ = 7.65 \times 10 ^ {-13} $ $ [H^+] = \sqrt{ K.C } = \sqrt { 7.65 \times 10 ^ {-13} \times 0.01 } $ $ = 0.87 \times 10 ^ { -7} M $ $ P ^ H = -log ( 8.7 \times 10^{-8} ) = 7.0605 $

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Solid $ Ba (NO_3)_2 $ is gradually dissolved in 1.0 $ \times 10 ^{-4} M Na_2CO_3 $ solution. At what concentration of will precipitate Ksp of $ BaCO_3 = 5.1 \times 10 ^ {-9} $

$ 4.1 \times 10 ^ {-5 } M $

$ 5.1 \times 10 ^ {-5} M $

$ 8.1 \times 10 ^ {-5} M $

$ 8.1 \times 10 ^ {-7} M $

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Explanation

$ K_{SP} of BaCo_3 = [Ba^{2+} ][CO^{2-} _3] $ $ \therefore [Ba ^{2+} ] = { 5.1 \times 10 ^ {-9} \over 1 \times 10 ^ {-4} } = 5.1 \times 10 ^ {-5} M $

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What is the $ [OH ^-] in the final solution prepared by mixing of 20.0 ml of 0.05 M HCl with 30.0 ml of 0.1 M Ba (OH)_2$ (?)

A. 0.4 M B. 0.05M C. 0.12 M D. 0.1 M

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Explanation

$ Ba(OH)_2 + 2HCl \rightarrow BaCl_2 + 2H_2O $ 2 mol HCl neutralize1 mole $ Ba(OH)_2 $ $ \therefore 1mol HCl neutralize0.5 mole Ba(OH)_2 $ $ Ba(OH)_2 \rightarrow Ba^{2+}+ 2OH ^ - $ 1 2 $ \therefore no.of molesof Ba(OH)_2 = 3 = 1 +2 $ $ \therefore Ba(OH)_2 left = 3 - 0.5 = 2.5 $ $ \therefore [ Ba(OH)_2 ] = { 2.5 \over 50 } = 0.05 M $ $ or [ OH^ - ] = 2 \times 0.05 M = 0.1 M $

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The ionisation constant of NH4OH is1.77 ï‚´10ï€5 at 298 k. Hydrolysis constant of it is

A. $ 6.5 \times 10 ^ {-12} $ B. $ 5.65 \times 10 ^ {-13} $ C. $ 5.65 \times 10 ^ {-12} $ D. $ 5.65 \times 10^{-10} $

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Explanation

$ K_h = { K_w \over K_b } = { 1.0 \times 10 ^ {-14} \over 1.7 \times 10^{-5} } = 5.65 \times 10 ^ {-10} $

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