Question

ame: Ka Hoa -8.5x100 (a Kw=1.000-14 Pla=-bella ) 38. Calculate the percent hydrolysis of the hypochlorite ion in 0.10 M NaOCI solution. a. 0.0012% PH= 7+1 plat - logo b. 0.024% 10.22 c. 0.056% PH= 7+ (1.457 2 108 10.10) = 4 d. 0.10% 10-11.0x10r 1474 10.22 - e. 0.17% Calculate the [OH-] in 0.20 M NaNO,. a. 4.8 x 10-9 M b. 1.4 x 10-8 M c. 1.2 x 10-7 M d. 6.5 x 10-7 M e. 2.1 x 10-6 M Not-thN Y 39.

Answer 1

OCl- + H2O -----> HOCl + OH-

kb = kw / ka = 10-14 / (3.5 x 10-8) = [HOCl][OH-]/ [OCl-]

now 2.857 x 10-7 = x2 / 0.10

x = 1.69 x 10-4

now percent of hydrolysis = (1.69 x 10-4 / 0.10) x 100 = 0.169%

= 0.17%

Answer is option (e)

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ka of HNO2 = 4.5 x 10-4

NO2- is base

find kb of NO2-

use

kb = 10-14 / ka

kb = 10-14 / (4.5 x 10-4)

kb = 2.22 x 10-11

NO2- dissociates as

NO2- + H2O -----> HNO2 + OH-

0.20 0 0

0.2 - x x x

kb = [HNO2][OH-] / [NO2-]

kb = x* x / (c-x)

assuming x can be ignored as compared to c

kb = x* x / (c)

x = sqrt(kb * c)

x = sqrt((2.22 x 10-11)*0.2)

x = 2.11 * 10-6 M

therfore [OH-] = 2.11 x 10-6 M

therefore answer is option(e)

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