Solve by ICE table
We dont know the amount of OH^- . Call it as x. The initial amount of C_2H_7N is 0.075M
color(white)(mmmmmmmm)C_2H_7N + "H"_2"O" ⇌ H_2N(CH_3)2^+ + OH^-color(white)(m)
"I/mol•L"^"-1":color(white)(mmml)0.075Mcolor(white)(mmmmmmm)0color(white)(mmmmml)0
"C/mol•L"^"-1":color(white)(mmmml)"-x"color(white)(mmmmmmmll)"+x"color(white)(mmmml)"+x"
"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
13.333333x^2=0.00059
Let's solve your equation step-by-step.
13.333333x^2=0.00059
Step 1: Divide both sides by 13.333333.
(13.333333x^2)/13.333333 = 0.00059/13.333333
x^2=0.000044
Step 2: Take square root.
x=sqrt0.000044
x=0.006652,
This is the OH^- concentration.
"pOH" = -log(OH^-)
-log(0.006652M) = 2.17704775945
"14 - pOH = pH"
14 - 2.17704775945 = 11.8229522406
