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How much sodium acetate should be added to 0.1 M solution $ CH_3COOH to give a solution of P^H 5.5 (pKa of CH_3CooH = 4.5 ) $
$ P^H = pka + { log [CH_3COONa] \over 1 [CH_3COOH] $ $ 5.5 = 4.5 + log { [CH_3COONa] \over 0.7 } $ $ 4.5 + log [CH_3COONa ] + 1 $ $ \therefore log [CH_3COONa] = 0 $ $ \therefore [CH_3COONa] = 7 M $
The $ P^H of solution obtained by mixing 50ml 0.4 N HCl \& 50ml 0.2 N NaOH $ is
$ 50 ml of 0.4 NHCl = { 0.4 \over 1000 } \times 50 = 0.02 g eq $ $ 50 ml of 0.2 N NaOH = { 0.2 \over 1000 } \times 50 = 0.01 g eq $ 0.01 g eq of NaOH will Neutalilise 0.01 g eq of HCl $ \therefore HCl left unneutralised = 0.01 g eq vol of Sol. =50+50 =100ml$ $ \therefore [HCl] = { 0.01 \over 100} \times 1000 = 0.1 N $ $ or [ H^+] = 0.1 M $ $ \therefore P ^ H = log (0.1) = 1.0 $
Ionisation constant of $ CH_3COOH is 1.7 \times10 ^{-5} and concentration of H ^+ ions is 3.4 \times 10 ^ {-4} The initial concentration of CH_3COOH$ molecules is
$ CH_3 COOH \rightleftharpoons CH_3 COO ^ - + H^+ $ $ at eq (a - 3.4 \times 10 ^ {-4} ) 3.4 \times 10 ^ {-4} 3.4 \times 10 ^ {-4} $ $ { (3.4 \times 10 ^ {-4} ) (3.4 \times 10 ^ {-4} ) \over ( a - 3.4 \times 10 ^ {-4} ) } = 1.7 \times 10 ^ {-5} (Given) $ $ \therefore a = 6.8 \times 10 ^ {-3} $
In the reversible reaction $ A + B \rightleftharpoons C + D $ , the concentration of each C and D at equilibrium was 0.8 mol litre, then the equilibrium constant $ K_C $ will be.
$ A+ B \rightleftarrow C + D $ initial 1 1 0 0 conc. Ateqm(1 - 0.8) 0.8 0.8 conc.(1- 0.8) = 0.2 = 0.2 $ Kc = { [C] [D] \over [A] [D] } = { 0.8 \times 0.8 \over 0.2 \times 0.2 } = 16.0 $
A reversible chemical reaction having two reactants in equilibrium. If the concentration of the reactants are doubled, then equilibrium constant will
$ K_C remains same beacause K_C $ is a characteristic constant.
Two moles of PCls are heated in a closed vessel of 2L capacity. At equilibrium, 40 % of $ PCl_5 is dissociated in to PCl_3 \& Cl_2 $ . The value of equilibrium constant is,
$ PCl_5 \rightleftharpoons PCl_3 + Cl_2 $ 2 0 0 $ {2 \times 60 \over 60 } { 2 \times 40 \over 100 } { 2 \times 40 \over 100} $ moles = 1.2 1.8 0.8 conc. = {mol \lit } = 1.2 / 2 0.8 / 2 0.8 /2 $ \therefore Kc = { [PCl_3 ] [Cl_2] \over [PCl_5] } = {0.4 \times 0.4 \over 0.6 } = 0.266 $
The dissociation constant for acetic acid and HCN at $ 25 ^\circ C are 1.5 \times 10 ^{-5} and 4.5 \times 10 ^{-10} respectively. the equilibrium constant for reaction CN^- + CH_3 COOH \rightleftharpoons HCN + CH_3COO ^ - $
$ Dissociation of CH_3 COOH CH_3 COOH \rightleftharpoons H ^ + + CH_3 COO ^ - K_{a1} = 1.5 \times 10 ^ {-3} $ $ Dissociation of HCN : HCN \rightleftharpoons H ^ = +CN^- k_{a2} = 4.5 \times 10 ^ {-3} $ For a reaction $ CN ^ - + CH_3 COOH \rightleftharpoons CH_3 COO ^ - + HCN is Ka = { Ka_1 \over Ka_2 } = { 1.5 \times 10 ^ {-3} \over 4.5 \times 10 ^ {-10} }= 3.33 \times 10 ^ 4 $
If in the reaction $ N_2O_4 \rightleftharpoons 2NO_2 , \alpha is that part of N_2O_4$ will dissociate, then the number of moles at equilibrium will be,
$ N_2 O_4 \rightleftarrow 2NO_2 $ 1 0 $ (1 - \alpha) $ $ 2 \alpha $ Total moles = $ 1- \alpha +2 \alpha =1+ \alpha from equation 2x=3 \therefore x = 3 /2 = 1.5 $
4.5 moles eachof hydrogen and iodine heated in a sealed ten litre vessel. At equilibrium, 3 moles of HI were found. The equilibrium constant for $ H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI(g) $ is
$ H_2 + I_2 \rightlleftarrow 2 HI $
Initial 4.5 4.5 0 conc. at eqm. (4.5-x)(4.5-x) 2x $ \therefore [H_2] = 4.5 -1.5 = 3 $ $ Kc = { [HI] ^2 \over [H_2 ] [I_2] } = { 3 \times 3 \over 3 \times 3 } = 1 $
The rate constant for forward and backward reaction of hydrolysis of ester are $ 1.1 \times 10 ^{-2 } \& 1.5 \times 10 ^ {-3 } $ per minute respectively. Equilibrium constant for reaction is, $ CH_3 COOC_2 H_5 + H_2 O \rightleftharpoons CH_3 COOH + C_2 H_5 OH $
$ K_f = 1.1 \times 10 ^ {-2} $ $ K_b = 1.5 \times 10 ^{-3} $ $ kc = { kf \over k_b } = { 1.1 \times 10 ^ {-2} \over 1.5 \times 10 ^ {-3} } = 7.33$
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