Question

Consider the titration of 42.5 mL of 0.193 M HCl with 0.125 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. (Assume that all solutions are at 25°C.)

Answer 1

Answer:

See explanation

Explanation:

Dilution law= C1V1=C2V2.

Where C is the concentration in mol per dm cube and V= volume in dm cube.

From the questions the parameters given are; V= 42.5ZmL, concentration or molarity of HCl= 0.193M and the molarity of KOH = 0.125M.

(a). At 0mL where the base KOH has not been added yet.

Moles of HCl = molarity × volume.

Moles of HCl = 0.193 × 42.5/1000

Moles of HCl= 0.00820205

pH=- log 0.0082025

pH= 2.086

Approximately, pH=2.1.

(b). At 3mL of 0.125M KOH

3/1000 × 0.125 mol/L

= 0.003L× 0.125 M

= 0.000375 moles KOH

Determining the number of moles of HCl, we have;

42.5/1000 × 0.193 mol/L

= 0.0082025

Final volume= 3mL+42.5mL= 45.5 ml

Final[HCl] = 0.0082025 - 0.000375 = 0.0078275.

Therefore, 0.0078275/ 0.0455

= 0.1720

pH= -log 0.1720

pH= 0.76.

(c). 5mL of 0.125M KOH

5/1000 × 0.125

= 0.005×0.125

= 0.000625 mol KOH

42.5/1000 × 0.193 mol/L

= 0.0082025 mol

Final volume= 5mL + 42.5= 47.5 mL.

0.0082025-0.000625= 0.0076

Final [HCl] = 0.0076/0.0475

= 0.16

pH= -log 0.16

pH= 0.79.

Answer 2

Answer:

When there is 0 mL of KOH added, the pH of the solution is 0.714

When there is added 3 mL of KOH, the pH of the solution is 0.764

When there is added 5 mL of KOH, the pH of the solution is 0.799

Explanation:

Step 1: Data given

Volume of 0.193 M HCl = 42.5 mL = 0.0425 L

Molarity of KOH = 0.125 M

Step 2: The balanced equation:

HCl + KOH --------> KCl + H2O

Step 3: pH of the solution when adding 0 mL of KOH

pH = -log[H+]

pH = -log (0.193)

pH = 0.714

Step 4: Calculate moles of HCl

Moles HCl = molarity * volume

Moles HCl = 0.193 M * 0.0425

Moles HCl = 0.0082 moles

Step 5: Adding 3 mL of KOH

When adding 3mL of KOH, the number of moles of KOH are:

Moles KOH = 0.125 M * 0.003 L

Moles KOH = 0.000375 moles

The mole ratio of the reacton is 1:1 so 0.000375 moles of KOH will react with 0.000375 moles of HCl

There will remain 0 moles of KOH

There will remain 0.0082 - 0.000375 = 0.007825 moles of HCl

The KCl formed, will not have an effect on the pH because it's a salt of a strong acid and strong base.

 

The final volume = 42.5 + 3 = 45.5 ml = 0.0455 L  

The new molarity of HCl = 0.007825/0.0455 = 0.172

HCl is a strong acid, this means it dissociates completely

HCl → H+ + Cl-

so [H+] = [HCl] = 0.172 M

pH = -log [H+] = -log(0.172) = 0.764

Step 6: pH when adding 5 mL of KOH

When adding 5 mL of KOH, the number of moles of KOH are:

Moles KOH = 0.125 M * 0.005 L

Moles KOH = 0.000625 moles

The mole ratio of the reacton is 1:1 so 0.000625 moles of KOH will react with 0.000625 moles of HCl

There will remain 0 moles of KOH

There will remain 0.0082 - 0.000625 = 0.007575 moles of HCl

The KCl formed, will not have an effect on the pH because it's a salt of a strong acid and strong base.

 

The final volume = 42.5 + 5 = 47.5 ml = 0.0475 L  

The new molarity of HCl = 0.007575/0.0475 = 0.159

HCl is a strong acid, this means it dissociates completely

HCl → H+ + Cl-

so [H+] = [HCl] = 0.159 M

pH = -log [H+] = -log(0.156) = 0.799