1.

pH of 0.1 M `NH_(3)` aqueous solution is `(K_(b) = 1.8 xx 10^(-5))`A. 11.13B. 12.5C. 13.42D. 11.55

Answer» Correct Answer - A
`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`.


Discussion

No Comment Found

Related InterviewSolutions