# Chemistry

A 25.0 mL sample of a 0.100 M solution of aqueous trimethylamine is titrated with a 0.125 M solution of HCl. What is the pH of the solution after 10.0, 20.0, and 30.0 mL of acid have been added?

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1. mola (CH3)3N initially = M x L = 0.100 x 0.025 = 0.00250 or 2.5 mmols.
mols HCl added @ 10 mL = M x L = 0.125 x 0.01 = 0.00125 or 1.25 mmols
mols HCl added @ 20 mL = 0.125 x 0.020 = 0.00250 or 2.50 mmols
mols HCl added @ 30 mL = 0.125 x 0.030 = 0.00375 or 3.75 mmols

You must know where each of these additions falls on the titration curve and the first thing you need to know is where is the equivalence point. mLa x Ma = mLb x Mb
mLa x 0.125 = 25 x 0.1 so mLa = 20 mL which tells you that 20 mL is the eq. pt, 10 is before and 30 is after the eq. pt.

................(CH3)3N + HCl ==> (CH3)3NH^+ + Cl^-
I.................2.5........0................0..................0
C...............-1.25..-1.25...........+1.25
E................1.25.......0..............1.25
You should recognize this as a buffered solution (a weak base and it's salt) so use the Henderson-Hasselbalch equation to determine the pH. I'll leave that for you.

At 20 mL that is th eq. pt. You have no (CH3)3N remaining and all is the salt ; i.e.(CH3)3NH^+. The pH is determined by the hydrolysis of the salt; i.e.,
..................(CH3)3NH^+ + H2O==> (CH3)3N + H3O^+
I..................0.0555..............................0...................0
C...................-x...................................x.....................x
E...................0.555-x..........................x.....................x
(NOTE)--0.0555 is the molar concn of the salt. You had 2.5 mmols salt and you have a total volume of 25 mL for the amine and another 20 added HCl to make 45 mL and 2.5/45 = 0.0555M
Ka for [(CH3)3NH^+] = Kw/Kb for [(CH3)3N] = ;(x)(x)/(0.0555-x)]
Substitute and solve for x = (H3O^+) and convert that to pH.

The 30 mL is after the eq. pt. There you calculate the mmols HCl used to get to the eq. Pt. subtract t find mmols HCl in excess, the (HCl) in excess is mmols in excess/mL total volume and convert to pH.

Post your work if you get stuck.

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2. Dr Bob, After addition of 10 ml of the HCl, this looks like a 50/50 buffered system in which [OH] = Kb = 4.4 x 10^-4, giving pOH = 3.36 and pH = 10.64. Would you be so kind as to check this. Thanks, Doc 48

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3. Based on the calculations that I see from Dr. Bob222, the pOH should be equivalent to the pKb after addition of 10mL, which is at the half equivalence point.

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4. That's what I'm showing in terms of pOH = pKb too. At half equivalence point the concentration of salt = concentration of amine and cancels from the Kb expression leaving Kb = [OH] => pKb = pOH.

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5. I agree with your assessment.

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6. Thanks to the two of you. I agree with you that the solution is a 50/50 buffered solution so that pH = pKa or pOH = pKb but I never calculated the pH or pOH value. I set up the ICE chart showing it was a 50/50 buffered solution and left it to the student to calculate the pH. And for whatever it's worth the question was to calculate the pH and not the pOH or Kb. I've tried for a while to understand. Are both of you saying I did something wrong? If so please spell it out. It appers to me, if I calculate the pH instead of just setting up the problem I get the same numbers both of you get. The pH I get is 10.64. Sorry if I'm thick headed. but I don't see a discrepancy. I think we have three people that are correct. Am I missing something? Thanks

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