Question #08525

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Question #08525

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Answer 1 · 2017-03-17T15:39:49

Solve by ICE table

We dont know the amount of $O H^{-}$ $OH^-$ . Call it as x. The initial amount of $C_{2} H_{7} N$ $C_2H_7N$ is 0.075M

$m m m m m m m m C_{2} H_{7} N + H_{2} O ⇌ H_{2} N (C H_{3}) 2^{+} + O H^{- m}$ $color(white)(mmmmmmmm)C_2H_7N + "H"_2"O" ⇌ H_2N(CH_3)2^+ + OH^-color(white)(m)$
${I/mol∙L}^{-1} : m m m l 0.075 M m m m m m m m 0 m m m m m l 0$ $"I/mol•L"^"-1":color(white)(mmml)0.075Mcolor(white)(mmmmmmm)0color(white)(mmmmml)0$
${C/mol∙L}^{-1} : m m m m l -x m m m m m m m l l +x m m m m l +x$ $"C/mol•L"^"-1":color(white)(mmmml)"-x"color(white)(mmmmmmmll)"+x"color(white)(mmmml)"+x"$
${E/mol∙L}^{-1} : m m m 0.075 M - x m m m m m l x m m m m m l l x$ $"E/mol•L"^"-1":color(white)(mmm) 0.075M"- x"color(white)(mmmmml)"x"color(white)(mmmmmll)"x"$

$K_{b}$ $K_b$ = $\frac{(H_{2} N (C H_{3}) 2^{+}) (O H^{-})}{C_{2} H_{7} N - x}$ ${( H_2N(CH_3)2^+) ( OH^-)} /(C_2H_7N -x$

$K_{b} = \frac{x^{2}}{0.075 - x}$ $K_b = x^2/(0.075-x$

Ignore x

$K_{b} = \frac{x^{2}}{0.075}$ $K_b = x^2/(0.075$

Solve for $x$ $x$

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

Let's solve your equation step-by-step.

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

Step 1: Divide both sides by 13.333333.

$\frac{13.333333 x^{2}}{13.333333} = \frac{0.00059}{13.333333}$ $(13.333333x^2)/13.333333 = 0.00059/13.333333$

$x^{2} = 0.000044$ $x^2=0.000044$

Step 2: Take square root.

$x = \sqrt{0.000044}$ $x=sqrt0.000044$

$x = 0.006652,$ $x=0.006652,$

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

$pOH = - log (O H^{-})$ $"pOH" = -log(OH^-)$

$- log (0.006652 M) = 2.17704775945$ $-log(0.006652M) = 2.17704775945$

$14 - pOH = pH$ $"14 - pOH = pH"$

14 - 2.17704775945 = 11.8229522406

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Question #08525

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