Answer:
1.14
Step-by-step explanation:
This is a neutralization reaction between KOH and HClO4, which can be represented by the following balanced chemical equation:
KOH + HClO4 → KClO4 + H2O
Before any HClO4 is added, the 20.0-mL sample of 0.150 M KOH contains:
moles of KOH = M × V = 0.150 mol/L × 0.0200 L = 0.00300 mol
Since the balanced chemical equation has a 1:1 stoichiometric ratio between KOH and HClO4, 0.00300 moles of HClO4 will be required to completely react with the KOH.
When 27.0 mL of 0.125 M HClO4 is added, the total volume becomes:
V_total = V_initial + V_added = 20.0 mL + 27.0 mL = 47.0 mL
The moles of HClO4 added can be calculated as:
moles of HClO4 = M × V = 0.125 mol/L × 0.0270 L = 0.00338 mol
Since the reaction is 1:1, this means that 0.00338 mol of KOH has been neutralized.
The moles of KOH remaining can be calculated as:
moles of KOH remaining = 0.00300 mol - 0.00338 mol = -0.00038 mol
This negative value indicates that all of the KOH has been neutralized and there is an excess of HClO4. Therefore, we need to calculate the pH of a solution of HClO4.
The concentration of HClO4 after 27.0 mL has been added can be calculated as:
M = moles of HClO4 / V_total = 0.00338 mol / 0.0470 L = 0.0719 M
The pH of this solution can be calculated using the equation:
pH = -log[H+]
Where [H+] is the concentration of hydrogen ions in the solution.
HClO4 is a strong acid, which means that it completely dissociates in water to form H+ and ClO4- ions. Therefore, the concentration of hydrogen ions in a 0.0719 M solution of HClO4 is equal to the concentration of HClO4:
[H+] = 0.0719 M
The pH can then be calculated as:
pH = -log[H+] = -log(0.0719) = 1.14
Therefore, the pH after 27.0 mL of 0.125 M HClO4 has been added to a 20.0-mL sample of 0.150 M KOH is 1.14.