Question

$\left[H^{+}\right]$ and $\left[O{H}^{-}\right]$ ion in a solution containing $9.25gm/L$ of $Ca(OH{)}_2$ at ${25}^{0}C$, is:

• A. $\left[O{H}^{-}\right]=0.25\times {10}^{-7},\left[H^{+}\right]=4\times {10}^{-7}$
• B. $\left[O{H}^{-}\right]=0.25,\left[H^{+}\right]=4\times {10}^{-14}$
• C. $\left[O{H}^{-}\right]=1.25\times {10}^{-7},\left[H^{+}\right]=8\times {10}^{-8}$
• D. $\left[O{H}^{-}\right]=2.5,\left[H^{+}\right]=4\times {10}^{-15}$

Answer 1

Answer: (B)

$\left[O{H}^{-}\right]=0.25,\left[H^{+}\right]=4\times {10}^{-14}$

$Ca(OH{)}_2=9.25gm/L$

Molar mass$=40+34=74$

Molarity$=\frac{9.25}{74}mol/L=0.125M$

$Ca(OH{)}_2\rightleftharpoons C{a}^{+2}+2O{H}^{-}$

$x$ $2x$

$\left[O{H}^{-}\right]=2\times 0.125=0.25M$

Now, we have,

$\left[H^{+}\right]\left[O{H}^{-}\right]={10}^{-14}$

$0.25\times \left[H^{+}\right]={10}^{-14}$

$\left[H^{+}\right]=4\times {10}^{-14}$