a) After adding 10 mL of HCl
first, we need to get moles of (CH3)3N = molarity * volume
= 0.29 m * 0.025 L
= 0.00725M moles
then, we need to get moles of HCl = molarity * volume
= 0.3625 m * 0.01L
= 0.003625 moles
so moles of (CH3)3N remaining = moles of (CH3)3N - moles of HCl
= 0.00725 - 0.003625
= 0.003625 moles
and when the total volume = 0.01 L + 0.025L = 0.035 L
∴ [(CH3)3N] = moles remaining / total volume
= 0.003625 moles / 0.035L
= 0.104 M
when we have Pkb so we can get Kb :
pKb = - ㏒Kb
4.19 = -㏒Kb
∴Kb = 6.5 x 10^-5
when Kb = [(CH3)3NH+] [OH-] / [(CH3)3N]
and by using ICE table we assume we have:
[(CH3)3NH+] = X & [OH] = X
and [(CH3)3N] = 0.104 -X
by substitution:
∴ 6.5 x 10^-5 = X^2 / (0.104-X) by solving for X
∴X = 0.00257 M
∴[OH-] = X = 0.00257 M
∴POH = -㏒[OH]
= -㏒0.00257
= 2.5
∴ PH = 14 - POH
= 14 - 2.5
= 11.5
b) after adding 20ML of HCL:
moles of HCl = molarity * volume
= 0.3625 m * 0.02 L
= 0.00725 moles
the complete neutralizes of (CH3)3N we make 0.003625 moles of (CH3)3NH+ So, now we need the Ka value of (CH3)3NH+:
and when the total volume = 0.02L + 0.025 = 0.045L
∴ [ (CH3)3NH+] = moles / total volume
= 0.003625 / 0.045L
= 0.08 M
when Ka = Kw / Kb
and we have Kb = 6.5 x 10^-5 & we know that Kw = 1 x 10^-14
so, by substitution:
Ka = (1 x 10^-14) / (6.5 x 10^-5)
= 1.5 x 10^-10
when Ka expression = [(CH3)3N][H+] / [(CH3)3NH+]
by substitution:
∴ 1.5 x 10^-10 = X^2 / (0.08 - X) by solving for X
∴X = 3.5 x 10^-6 M
∴ [H+]= X = 3.5 x 10^-6 M
∴PH = -㏒[H+]
= -㏒(3.5 x 10^-6)
= 5.5
C) after adding 30ML of HCl:
moles of HCl = molarity * volume
= 0.3625m * 0.03L
= 0.011 moles
and when moles of (CH3)3N neutralized = 0.003625 moles
∴ moles of HCl remaining = moles HCl - moles (CH3)3N neutralized
= 0.011moles - 0.003625moles
= 0.007375 moles
when total volume = 0.025L + 0.03L
= 0.055L
∴[H+] = moles / total volume
= 0.007375 mol / 0.055L
= 0.134 M
∴PH = -㏒[H+]
= -㏒ 0.134
= 0.87
