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$ In 2 HI \rightleftharpoons H_2 + I_2 \triangle H \gt O $ the forward reaction is affected by change in
For endothermic $ ( \triangle H \gt O ) $ reaction, change in temperature affects the equilibrium system and forward reaction takes place . by increasing temp .
If $ K_1 and K_2 are respective equlibrium constants for two reaction, XeF_{6(g)} + H_2O_{(g)} \rightleftharpoons XeOF_{4(g) } + 2HF_{(g)} XeF_{4(g)} + XeF_6 \rightleftharpoons XeOF_{4(g)} + XeO_3F_{2(g)} the equilibrium constant for the reaction XeF_{4(g)}+ 2HF_{(g)} \rightleftharpoons XeO_3F_{2(g)}+ H_2O_{(g)} $ will be
$ Kc = { [ NO] ^4 [ H_2 O ] ^ 6 \over [ NH_3 ] ^4 [ O_2 ] ^5 } = (conc.) ^ { 4+6-(4+5)} $ $ = conc ^ {-1} $
For a homologous reaction, $ 4 NH_3 + 5O_2 \rightleftharpoons 4 NO + 6H_2O $ the dimensions of equilibrium constant $ K_C $ is
According to Le-chatelliers principle, if conc. of reactant become doubled, then forward reaction takes place and concentration of product also increases. so equilibrium constant also remains same.
$P^H$ of 0.005 M calcium acetate $( P^{Ka}$ of $CH_3COOH$ = 4.74 ) is
change in volume affects number ofmoks per unit volume. In a reaction
One of the following equilibria is not affected by change in volume of the flask.
$ N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)} $ no. of moles of reactants and product are equal so volume change does not affects the equilibrium
$ P^H of 10 ^ {-8 }$ M solution of Hclin water is
$ P^H of HCl should be less than 7. due to self ionisation of H_2O $ from acid $ [H ^+] = 10 ^ {-8} M from H_2O [H ^+] = 10^{-7 } M $ $ Total [H ^+] = 10 ^ {-8} + 10^{-7} = 10 ^{-8 } (1 + 10) = 11 \times 10 ^ {-8} M $ $ P^H = - log [H^+] = - log(11 \times 10^{-8}) = - (1.0414 - 8) = 6.96 $ $P ^H =6.96 $
A certain buffer solution contains equal concentration of $ X ^ - and HX . Ka for HX is 10^{-8} . The $ P^H $ of buffer is
$ P^H $ of buffer solution $ = p K_a + log { [ salt ] \over [ Acid] } $ $ = - log K_a + log 1 $ $ = - log 10 ^ {-8} + 0 = +8 $
The solubility product of AgCl is $ 4 \times 10^{-10} at 298 k The solubility of AgCl in 0.04M CaCl_2 $ willbe
$ if x is the solubility of AgCl in 0.04 M cacl2 , then $ [Ag ^ + ] = x mol L ^ {-1} $ $ [Cl ^ - ] = 2 \times 0.04 + x = 0.08 + x \cong 0.08 M $ $ K_{SP} of AgCl = [Ag ^ + ] [Cl ^ - ] $ $ 4 \times 10 ^ {- 10} / 0.08 = [Ag ^ + ] = 5 \times 10^ {-9} M $
Calculate concentration ofsodium acetate which should be added to 0.1 M solution of $ CH_3 COOH (P^{K_a} = 4.5 ) $ to give a solution of $ P^H 5.5 $
$ P ^ H = pka + log { [ CH_3 COONa] \over [ CH_3 COOH] } $ $ 5.5 = 4.5 + log { [ CH_3 COONa ] \over 0.1 } $ $ 5.5 = 4.5 + log [ CH_3 COONa] + 1 $ $ \therefore log [ CH_3 COONa ] = 0 $ $ \therefore [ CH_3 CooNa] = 1 M $
Which of the following is a base according to lowry-bronsted concept ?
$ I ^ - $ can accept protons and hence is a base.
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