Respuesta :
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:
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
