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
