Answer to Question #153010 in General Chemistry for Joy

Question

A 0.050 M concentration of weak base, CH3CH2NH2, was found to ionize at 1.7%. What is the
concentration of CH3CH2NH3⁺ in the solution?

Answer 1

$CH_3CH_2NH_{2(aq)} + H_2O_{(l)} ⇌ CH_3CH_2NH_{3(aq)}^+ + OH^-_{(aq)}$

$[CH_3CH_2NH_{2}] = 0.050M$

ionization% = $\dfrac{x}{[CH_3CH_2NH_{2}]} × 100\%$

x = [OH^-] = $\dfrac{1.7\% × 0.050M }{100\%}$ = 8.5 × 10-⁴ M

because the reaction is in equilibrium,

$[CH_3CH_2NH_{3(aq)}^+] = [OH^-]$

$\therefore [CH_3CH_2NH_{3(aq)}^+] = 8.5 × 10^{-4} M$

In conclusion, the concentration of CH₃CH₂NH₃⁺ in the solution is 8.5 × 10-⁴ M.