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Solution Equilibrium Problem?
Please Help. Thanks in advance!
40ml of 0.35M CH3NH2 is titrated with 0.28M HCl until the end point. Calculate the pH at endpoint. Kb = 5 * 10^-4.
1 Answer
- BobbyLv 75 years agoFavorite Answer
CH3NH2 + H3O+ = CH3NH3+ + H2O
volume of HCl = 50.0mL
at the end point we have CH3NH3+ in 90.0 ml of water
moles of CH3NH3+ = 0.35 M * .040 = 0.014 mol = 0.014 moles / .090 l = 0.155M
at the eq point we have
CH3NH3+(aq) + H2O(l) = CH3NH2(aq) + H3O+(aq)
Ka= Kw/Kb = 10 ^-14 / 5 * 10 ^-4 = 2 *10^-11
at the eq point x moles of H3O + and CH3NH2 are formed and 0.155 - x moles of CH3NH3+ are left
Ka = [CH3NH2(aq)] * [H3O+(aq)] / [CH3NH3+(aq)]
2 *10^-11 = x^2 / (0.155 - x)
x is very small compared to 0.155 so
x^2= 0.155 * 2 *10^-11
x= 1.760 *10^-06 = [H3O+]
pH = 5.76
