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
<em>Step 3: pH of the solution when adding 0 mL of KOH</em>
pH = -log[H+]
pH = -log (0.193)
<u>pH = 0.714</u>
Step 4: Calculate moles of HCl
Moles HCl = molarity * volume
Moles HCl = 0.193 M * 0.0425
Moles HCl = 0.0082 moles
<em>Step 5: Adding 3 mL of KOH </em>
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) = <u>0.764</u>
<em>Step 6: pH when adding 5 mL of KOH</em>
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) = <u>0.799</u>