Answer in General Chemistry for Joy #153010
March 17th, 2023
“CH_3CH_2NH_{2(aq)} + H_2O_{(l)} u21cc CH_3CH_2NH_{3(aq)}^+ + OH^-_{(aq)}”
“[CH_3CH_2NH_{2}] = 0.050M”
ionization% = “dfrac{x}{[CH_3CH_2NH_{2}]} u00d7 100%”
x = [OH–] = “dfrac{1.7% u00d7 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 u00d7 10^{-4} M”
In conclusion, the concentration of CH3CH2NH3+ in the solution is 8.5 × 10-⁴ M.