Question

3.3 x 10-1 M 15.0 mL of 0.50 M HCl is added to a 100.0 ml sample of 0.377 M HNO 2(Ko for HNO 2 - 4.0 x 104). What is the equilibrium concentration of NO 2-ions? 1.3 x 10-4 M 4.5 x 10-2M 2.0 x 10-3M none of these

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Answer 1

to Let us calculate the final concentration of each in the new solution by adding 15.0 mL of 0.50m Ha 100.0 mL of sample of 0.377 M HNO2. 0.377 M X 100.0 mL (100.0+ 15.0) m findl concentration of HNO₂ -0.327 826087 M As, 1 molecute of HNO2 disscriates to give a molecule of tt and No ions. Let, at equilibrium, each 'x molar HNO2 has dissociated so, HNO H+ + Nog at time=0, 0.3278 M O M .OM at t= teg (0.3271-2) M х М х М so, ka= [H + { [ 02 Jug - 4,0x10-4 [ HNO2 Jeg As, concentration of ttion will be much less, is, x, is much less (as dissociation constant is very small)it con be assumed that (0.327-280.3278 for simplification. x So, ka = 410 410-4 0.3278 => x=80.3278 x 40 x 10 - 0.01145 M

So in final solution, the concentration of [+] ion coming from HNO will be = x=0.1145 M Now as Hal На is a strong acid, it will undergo complete dissociation and the final concentration of will be the final concentration of itt ions coming from Ha. So, [A+] from Ha 0.50mx 15.0 mL (15.0+ 100.0) m = 0.0652 M ML Nao, the total concentration of Htion in the solution = [H + I total = + 0.0652) = (0.0145 0.07665M Now, putting these values in equation 0 y

4.04 10-4 [0.0766 ] [ wozu [0.3278] to keep ka value intact, as [+] has increased, [no] must decrease to maintain the constant ka. 7 [Nog (4.0 x10-4) x (0-3270) (0.0766) M. [NO2] = 1.7117 x 10-3 M None of these