1.

pH of 0.1 M NH_(3) aqueous solution is(K_(b) = 1.8 xx 10^(-5))

Answer»

11.13
12.5
13.42
11.55

Solution :`NH_(4)OH HARR NH_(4)^(+) + OH^(-)`
`K_(b) = C alpha^(2) , (1.8 xx 10^(-5))/(.1) = alpha^(2), alpha = 1.34 xx 10^(-3)`
`[OH^(-)] = alpha. C = 1.34 xx 10^(-3) xx .1`
`POH = LOG 10(1)/(1.34 xx 10^(-4)) , pOH = 2.87`
`pH = pOH = 14 , pH + 2.87 = 14`
`pH = 14 - 2.87, pH = 11.13`.


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