# Question 08525

Mar 17, 2017

Solve by ICE table

We dont know the amount of $O {H}^{-}$ . Call it as x. The initial amount of ${C}_{2} {H}_{7} N$ is 0.075M

color(white)(mmmmmmmm)C_2H_7N + "H"_2"O" ⇌ H_2N(CH_3)2^+ + OH^-color(white)(m) 
$\text{I/mol•L"^"-1} : \textcolor{w h i t e}{m m m l} 0.075 M \textcolor{w h i t e}{m m m m m m m} 0 \textcolor{w h i t e}{m m m m m l} 0$
$\text{C/mol•L"^"-1":color(white)(mmmml)"-x"color(white)(mmmmmmmll)"+x"color(white)(mmmml)"+x}$
$\text{E/mol•L"^"-1":color(white)(mmm) 0.075M"- x"color(white)(mmmmml)"x"color(white)(mmmmmll)"x}$

${K}_{b}$ ={( H_2N(CH_3)2^+) ( OH^-)} /(C_2H_7N -x

K_b = x^2/(0.075-x

Ignore x

K_b = x^2/(0.075#

Solve for $x$

$13.333333 {x}^{2} = 0.00059$

$13.333333 {x}^{2} = 0.00059$

Step 1: Divide both sides by 13.333333.

$\frac{13.333333 {x}^{2}}{13.333333} = \frac{0.00059}{13.333333}$

${x}^{2} = 0.000044$

Step 2: Take square root.

$x = \sqrt{0.000044}$

$x = 0.006652 ,$

This is the $O {H}^{-}$ concentration.

$\text{pOH} = - \log \left(O {H}^{-}\right)$

$- \log \left(0.006652 M\right) = 2.17704775945$

$\text{14 - pOH = pH}$

14 - 2.17704775945 = 11.8229522406