Given the following data, Base Ko NH: 1.8 x 10-5 CH3NH2 5.0 x 10-4 C6H6NH, 4.2...

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Here all the compounis given are salts of weak base and stroong acid. Given concentration of all the salts=0.52m for this type of salts pH can be calcu- lated by the equation - pH = = - 2 pko-ž loge £ NHA NOz - Salt of weak base NH3 - and stroong acid HNO₃ conc. of salt, c = 0.52 M Given, ko for NH3, ko = 1.8x10-5 i log kb = log (1.8*10-5) a - log ko = -log (1.8 x 10-5) a pkb = 4.74 Now using eqn @ we have pH = 7-12 (4.74) - log (0.52) = 7- 2.37 +0.14 we

a pH = 4.77 ~ 4.8 two significant figures ли CH3NH₂ NOz : weak base =) CHg NHg a strong acid =) HNO₂ Given ko = 5.0 x10-4 .. pkp =- log (5.0*10-4) = 3.3 Cone, c = 0.52 M froom equation ③ we have a pH = 7 - 1 (3.3) -12 log (0.52) Ne Yo02 a on = 7 - 1:65 + 0.14 = 5.49 Ž 5.5 two significant figures Thus, pH = 15.5 Aus - Cale NH3 NO:) weak base C6H5NH2 Strong acid — HNO3

Giren, ko = 4.2x 10-10 pkb = -log (4. 2x10-10) = 9.37 Given, conc, c = 0.52M from equation 6 - pH = 7-1 (9.37) - I log (0.52) = 7-4.68 +0.14 = 2.46 ²2.5 two significant figures ] Thus, pH = 2.5 Aus