Question

What are the [H₃O⁺] and pH of 0.085M weak base C₂H₅NH₂?

Answer 1

$K_b=6.4\times10^{-4}$

$K_b=$ $[C_2H_5NH_3^{+}][H_3O^+]\over[C_2H_5NH_2]$

$[0+x][0+x]\over[0.085-x]$

$K_b=$ $[x][x]\over [0.085M]$

$6.4\times 10^{-4}=$ $[x^2]\over [0.085]$

$6.4\times 10^{-4}[0.085]=x^2$

$\sqrt{(6.4\times 10^{-4}}[0.085]$ $=\sqrt{x^2}$

$H_3O^{+}=0.00783M$

$pOH=-log[0.00738]$

$pOH=2.132$

$pH+pOH=14$

$pH=14-2.132=11.87$