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An example for lewis acid is
Lewis acid always accepts a pair of $ e ^ - here AlCl_3 accepts a pair of e ^ - $ .
Which of the following molecule act as a lewis acid ?
$ (CH_3)_3 B accept a pair of e ^ - $ so lewis acid
In a given system, water and ice are in equilibrium. If pressure is applied to the above system, then,
Pressure on equilibrium system increases ,so volume decreases. volume
of ice is more than liquid $ H_2O$ so more ice is malted
$ 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 $
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