What is the hydronium ion concentration of a 0.02M solution of Cu²⁺ solution of copper (II) perchlorate? The acidity constant of the following reaction is 5×10^–9

$\large C{u^{2 + }}_{\left( {aq} \right)}\, + \,2{H_2}{O_{(l)}}\, \rightleftharpoons \,Cu(OH)_{(aq)}^ + \, + \,{H_3}O_{(aq)}^ +$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

5×10^–4

7×10^–4

1×10^–5

1×10^–4

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$\mathop {Cu{{\left( {Cl{O_4}} \right)}_2}}\limits_{{C_0}} \xrightarrow{{}}\mathop {C{u^{ + 2}}}\limits_{{C_0}} + \mathop {2ClO_4^ - }\limits_{{C_0}}$

$\mathop {C{u^{ + 2}}}\limits_{{C_0} - {C_0}\alpha } + 2{H_2}O \rightleftharpoons \mathop {{{\left[ {Cu\left( {OH} \right)} \right]}^ + }}\limits_{{C_0}\alpha } + \mathop {{H_3}{O^ + }}\limits_{{C_0}\alpha }$

${K_h} = {K_a} = 5 \times {10^{ - 9}} = \frac{{{C_0}{\alpha ^2}}}{{1 - \alpha }}$

neglecting α we have,

$5 \times {10^{ - 9}} = {C_0}{\alpha ^2}$

$\Rightarrow {C_0} \times 5 \times {10^{ - 9}} = {C_0}\left( {{C_0}{\alpha ^2}} \right) = {\left( {{C_0}\alpha } \right)^2} = {\left[ {{H_3}{O^ + }} \right]^2}$