A 25.0 ml sample of a 0.2900 m solution of aqueous trimethylamine is titrated with a 0.3625 m solution of hcl. calculate the ph of the solution after 10.0, 20.0, and 30.0 ml of acid have been added; pkb of (ch3) 3n = 4.19 at 25°c.

Question

Jakob Whitney

Chemistry

23 February, 17:52

A 25.0 ml sample of a 0.2900 m solution of aqueous trimethylamine is titrated with a 0.3625 m solution of hcl. calculate the ph of the solution after 10.0, 20.0, and 30.0 ml of acid have been added; pkb of (ch3) 3n = 4.19 at 25°c.

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Sullivan Velazquez · Answer 1

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