Homework Help: This question has multiple parts A titration enthusiast titrates

Question

Homework Help: This question has multiple parts

A titration enthusiast titrates because he wants to titrate things because that is what good scientist do...they titrate.

Part A: *I Tried This One. I'm sure I'm wrong.* Titration of 50mL of 0.125M NaOH is added to a 50mL solution of 0.25M formic acid (HCOOH). Ka(HCOOH) = 1.7E-4. What is the pH after the addition of NaOH?

My Answer: pH=2.60 which is too low for this.

Part B: How many mL of 0.125M NaOH need to be added to the 50mL solution of 0.25M formic acid (HCOOH) to reach the equivalence point? What is the pH at the equivalence point? Ka is the same as part A.

Part C: Enthusiast titrates again. This time, a 100 mL solution of 0.13M trimethylamine ((CH3)3N) is titrated with 25mL of 0.31M HCl. Kb ((CH3)3N) =6.4E-5. What is the pH of the solution AFTER the addition of the HCl?

Part D: How many mL of 0.31M HCl need to be added to the 100mL solution of the 0.13M trimethylamine ((CH3)3N) to reach the equivalence point? What is the pH at the equivalence point? Kb is the same as in Part C.

Explanation / Answer

the solution is as follows

Part A: Titration of 50 mL of 0.125 M NaOH is added to a 50 mL solution of 0.25 M formic acid (HCOOH). (Ka(HCOOH) = 1.7E-4).

Equation:
HCOOH + NaOH ---> HCOONa + H2O
1 mol HCOOH reacts with 1 mol NaOH to produce 1 mol HCOONa plus water

Find pKa of HCOOH

pKa = -log (Ka)

       = -log (1.7 x 10^-4)

       = 3.77

You will notice that throughout most of the problem you have a solution containing a weak acid ( HCOOH) and the conjugate base of that acid , from the salt HCOONa. This is a buffer solution .

After addition of 50.0 mL of NaOH. This produces a mixture of unreacted HCOOH and HCOONa in solution . This is a buffer solution . You determine the pH of a buffer solution using the Henderson - Hasselbalch equation.

We will use the above working out , but shortened:
Mol HCOOH in 50.0mL of 0.25 M HCOOH solution = 50/1000*0.25

                                                                                  = 0.0125 moles HClO
Mol KOH in 50.0ml of 0.125 M solution = 50/1000*0.125

                                                              = 0.00625 mol KOH
These react to produce 0.00625 mol of HCOONa ,

and, there is 0.0125-0.00625 = 0.00625 mol unreacted HCOOH remaining.
Calculate the molarity of each compound in solution: Final volume = 50+50 = 100 mL = 0.1 L
HCOOH = 0.00625/0.1 = 0.0625 M
HCOONa = 0.00625/0.1 = 0.0625 M

Now apply the H-H equation:
pH = pKa + log ( [salt] /[acid])
pH = 3.77 + log ( 0.0625/0.0625)
pH = 3.77

Part B : To find mL of 0.125 M NaOH need to be added to the 50mL solution of 0.25 M formic acid (HCOOH) to reach the equivalence point and to calculate the pH at the equivalence point.

At equivalence point moles of acid remaining will be equal to the moles of salt in solution. Such that, the situation will be same as in Part A.

thus, mL of NaOH to be added = 50 mL

pH = pKa at equivalence point = 3.77

Part C : 100 mL solution of 0.13M trimethylamine ((CH3)3N) is titrated with 25mL of 0.31M HCl. Kb ((CH3)3N) =6.4E-5

1 mole of trimethylamine is neutralized with 1 mole of HCl

Kb for trimethyamine = 6.4 x 10^-5

moles of trimethylamine = M x L

                                    = 0.13 x 0.1

                                    = 0.013 moles

moles of HCl = 0.31 x 0.025

                     = 0.00775 moles

moles of salt formed = moles of acid added = 0.00775 moles

moles of trimethylamine remaining = 0.013 - 0.00775

                                                     = 0.00525 moles

Total volume of solution = 100 + 25 = 125 mL = 0.125 L

Molarity of trimethylamine = 0.00525 / 0.1 = 0.0525 M (remaining in solution)

Me3N + H2O ----> Et3NH+ + OH-

Kb = [Me3NH+][OH-]/[Me3N]

Let x be the concentration of [Me3NH+] = [OH-], then

6.4 x 10^-5 = (x)(x)/0.0525

x = [OH-] = 1.833 x 10^-3

pOH = -log(OH-) = -log(1.83 x 10^-3) = 2.74

pH = 14 - pOH = 14 - 2.74 = 11.26

Part D. mL of 0.31M HCl needed to be added to the 100mL solution of the 0.13M trimethylamine ((CH3)3N) to reach the equivalence point.

At equivalence point all the trimethylamine is present in salt form. [Salt] = [remaing acid]

Moles of trimethylamine = 0.13 M x 0.1 L = 0.013 moles

moles of salt in solution = 0.013 moles

Remaining acid should be = 0.013 moles

So we would be needing 0.013 x 2 moles of acid = 0.026 moles

thus,

L of 0.31 M of HCl required = 0.026/0.31 = 0.0839 L = 83.9 mL

Thus. 83.9 mL of HCl is needed to reach equivalence point

Now calculate pH at equivalence point

moles of acid remaining at ewuivalence point = 0.013 moles

Total volume = 83.9 + 100 mL = 183.9 mL = 0.1839 L

Molarity of HCl = 0.013/0.1839 = 0.071 M

HCl is a strong acid will dissociate completely

So [H+] = 0.071 M

pH = -log(0.071) = 1.15

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