Question

A 0.02M monobasic acid has a degree of dissociation 0.12. Its pH is
a) 5.4
b) 2.6
c) 2.4
d) 5.2

Answer 1

$\sf \bold {Given\::}$

$\bullet$ concentration of monobasic acid (c) = 0.02M

$\bullet$Degree of dissociation (α) = 0.12

$\sf \bold {To\:find\::}$ pH of solution

$\sf \bold {Formula\:used\::}$

$\bullet$ $\sf [H^{+}]$ = c × α

$\bullet$ pH = - log $\sf [H^{+}]$

$\sf \bold {Solution\::}$

$\sf [H^{+}]$ = cα = 0.02×0.12

$\sf [H^{+}]$ = 0.0024 = 2.4 × $\sf 10^{-3}$

pH = - log $\sf [H^{+}]$

pH = - log ( 2.4 × $\sf 10^{-3}$)

pH = - ( log 2.4 + log $\sf 10^{-3}$)

pH = - ( log 2.4 + (-3) log 10 )

pH = - ( 0.3802 - 3 )

pH = - ( -2.6198)

pH = 2.6198

$\sf \bold {ANSWER\::} \:option \:b)$ $\sf \bold {pH\:=\:2.6198\::}$