1) A 25.0 mL sample of a .290 M solution of aqueous trimethylamine is titrated w

Question

1) A 25.0 mL sample of a .290 M solution of aqueous trimethylamine is titrated with a .363 M solution of HCl. Calculate the pH of the solution

a) after 10.0 mL of acid have been added:

b) after 20.0 mL of acid have been added:

c) after 30.0 mL of acid have been added:

*pKb of (CH3)3N=4.19 at 25 degrees C.11

2) Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.00 mL sample of the cleaner, the equivalence point is reached after 16.81 mL of .135 M HCl has been added. What is the

a) initial concentration of ammonia in the solution?

b) pH of the solution at the equivalence point?

Explanation / Answer

1) A 25.0 mL sample of a .290 M solution of aqueous trimethylamine is titrated with a .363 M solution of HCl. Calculate the pH of the solution

a) after 10.0 mL of acid have been added:

solution :- balanced reaction equation

(CH3)3N + HCl ------ > (CH3)3NH+   + Cl-

*pKb of (CH3)3N=4.19 at 25 degrees

Lets first calculate the moles of the (CH3)3N and HCl using their molarity and volumes

Moles = molarity * volume in liter

Moles of (CH3)3N= 0.290 mol per L * 0.025 L = 0.00725 mol

Moles of HCl =0.363 mol per L * 0.010 L = 0.00363 mol

After the reaction (CH3)3N will react with HCl and forms (CH3)3NH+

So lets calculate the new moles of (CH3)3N and (CH3)3NH+

New moles of (CH3)3N = 0.00725 mol – 0.00363 mol = 0.00362 mol

New moles of (CH3)3NH+ = 0.00363 mol

Now lets calculate new molarities at the total volume (25 ml + 10 ml = 35 ml = 0.035 L)

New molarity of (CH3)3N =0.00362 mol / 0.035 L =0.1034 M

New molarity of (CH3)3NH+ = 0.00363 mol / 0.035 L =0.1037

Now lets calculate the pH using the Henderson equation

pOH = pkb + log ([acid]/[base])

lets put the values in the formula

pOH = 4.19 + log ([0.1037]/[0.1034])

pOH =4.19+0.001258

pOH = 4.19

pH= 14- pOH

pH= 14-4.19

pH =9.81

b) after 20.0 mL of acid have been added:

Solution

Lets calculate the moles of the (CH3)3N and HCl

Moles of (CH3)3N= 0.290 mol per L * 0.025 L = 0.00725 mol

Moles of HCl =0.363 mol per L * 0.020 L = 0.00726 mol

Moles of acid are more than moles of base therefore lets find the excess moles of acid remaining after the reaction

Moles of HCl remain after reaction = 0.00726 -0.00725 = 0.00001 mol

Now lets calculate its new molarity at total volume 25 ml + 20 ml = 45 ml = 0.045 L

New molarity of HCl = 0.00001 mol / 0.045 L =0.000222M

Now lets calculate pH using this concentration

pH = -log[H+]

pH= -log[0.000222]

pH =3.65

c) after 30.0 mL of acid have been added:

Solution :-

Lets calculate the moles of the (CH3)3N and HCl

Moles of (CH3)3N= 0.290 mol per L * 0.025 L = 0.00725 mol

Moles of HCl =0.363 mol per L * 0.030 L = 0.01089 mol

Moles of HCl are excess therefore lets calculate the excess moles of HCl remaining after reaction

Moles of HCl remain after reaction = 0.01089 mol – 0.00725 mol =0.00364 mol HCl

Now lets calculate new molarity of HCl at total volume 25 ml + 30 ml = 55 ml =0.055 L

New molarity of HCl =0.00364 mol / 0.055 L = 0.06618 M

Now lets calculate the pH

pH= -log[H+]

pH= -log[0.06618]

pH= 1.18

2) Your job is to determine the concentration of ammonia in a commercial window cleaner. In the titration of a 25.00 mL sample of the cleaner, the equivalence point is reached after 16.81 mL of .135 M HCl has been added. What is the

a) initial concentration of ammonia in the solution?

Solution

Balanced reaction equation is as follows

NH3 +HCl ----- > NH4+ + Cl-

Lets first calculate moles of HCl using its molarity and volume

Moles of HCl = 0.135 mol per L *0.01681 L =0.00227 mol HCl

Since mole ratio of the HCl and NH3 is 1 :1 therefore moles of NH3 reacted are same as moles of HCl

So moles of NH3 = 0.00227 mol

Now lets calculate molarity of the NH3

Molarity = moles / volume in liter

Initial Molarity of NH3 = 0.00227 mol / 0.025 L = 0.0908 M

Therefore initial concentration of the NH3 = 0.0908 M

b) pH of the solution at the equivalence point?

Solution :-

At the equivalence point all the NH3 will converted into the NH4+ ions

Lets calculate the new molarity of the NH4+ at the equivalence point when total volume is 25.0 ml + 16.81ml = 41.81 ml = 0.04181 L

[NH4+] = 0.00227 mol / 0.04181 L =0.05426 M

Therefore lets calculage the concentration of the H3O+ using the ice table and ka value of the NH4+

          NH4+      + H2O ----- > H3O+   + NH3

I     0.05429 M                            0              0

C    -x                                +x                        +x

E 0.05429-x                      x                          x    

Kb of NH3 = 1.8*10^-5

Therefore ka of NH4+ = kw/kb

Ka= 1*10^-14 /1.8*10^-5 = 5.56*10^-10

Now lets calculate the [H3O+]

Ka = [H3O+][NH3]/[NH4+]

5.56*10^-10 = [x][x]/[0.05429 –x]

Since ka is very small therefore we can neglect the x from the denominator then we get

5.56*10^-10 = [x][x]/[0.05429 ]

5.56*10^-10 *0.05429 = x^2

3.02*10^-11=x^2

By taking square root of both sides we get

5.5*10^-6 =x

Now lets calculate the pH using this concentration of H3O+

pH= -log[H3O+]

pH= -log[5.5*10^-6]

pH= 5.26

therefore pH at equivalence point = 5.26

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.