A 30 mL sample of 0.150 M KOH is titrated with 0.125 M HClO4 solution. Calculate the pH after the following volumes of acid have been added: 30 mL, 35 mL, 36 mL, 37 mL, and 40 mL.
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To calculate the pH after the given volumes of acid have been added, we need to determine the moles of acid and base present in the solution at each step. Then, we can use the balanced chemical equation and the Henderson-Hasselbalch equation to find the pH.
Step 1: Calculate the number of moles of KOH initially present:
moles of KOH = volume of KOH solution (in L) x concentration of KOH
moles of KOH = 0.030 L x 0.150 M
moles of KOH = 0.0045 mol
Step 2: Calculate the number of moles of HClO4 added:
moles of HClO4 = volume of HClO4 solution (in L) x concentration of HClO4
a) After 30 mL of HClO4 added:
moles of HClO4 = 0.030 L x 0.125 M
moles of HClO4 = 0.00375 mol
Step 3: Determine the limiting reactant:
The limiting reactant is the one that is completely consumed in the reaction. In this case, KOH and HClO4 react in a 1:1 mole ratio, so the limiting reactant is determined by the smaller number of moles.
Since 0.00375 mol of HClO4 was added and the initial moles of KOH is 0.0045 mol, HClO4 is the limiting reactant.
Step 4: Determine the moles of excess reactant:
Excess reactant refers to the reactant that is not completely consumed. In this case, KOH is the excess reactant.
moles of excess KOH = initial moles of KOH - moles used (moles of HClO4)
moles of excess KOH = 0.0045 mol - 0.00375 mol
moles of excess KOH = 0.00075 mol
Step 5: Calculate the moles of acid remaining:
moles of acid remaining = moles of excess KOH
moles of acid remaining = 0.00075 mol
Step 6: Calculate the concentration of the acid remaining:
concentration of acid remaining = moles of acid remaining / volume of solution remaining
concentration of acid remaining = 0.00075 mol / (30 mL - volume of acid added)
a) After 30 mL of HClO4 added:
concentration of acid remaining = 0.00075 mol / (30 mL - 30 mL) = 0.00075 mol / 0 mL (undefined)
b) After 35 mL of HClO4 added:
concentration of acid remaining = 0.00075 mol / (30 mL - 35 mL) = 0.00075 mol / -5 mL (undefined)
c) After 36 mL of HClO4 added:
concentration of acid remaining = 0.00075 mol / (30 mL - 36 mL) = 0.00075 mol / -6 mL (undefined)
d) After 37 mL of HClO4 added:
concentration of acid remaining = 0.00075 mol / (30 mL - 37 mL) = 0.00075 mol / -7 mL (undefined)
e) After 40 mL of HClO4 added:
concentration of acid remaining = 0.00075 mol / (30 mL - 40 mL) = 0.00075 mol / -10 mL (undefined)
Since the volume of the solution remaining is negative or zero, it is not possible to calculate the concentration of the acid remaining and therefore the pH cannot be determined for these volumes.
To calculate the pH after each volume of acid is added, we need to consider the reaction between KOH and HClO4. KOH is a strong base that dissociates completely in water, while HClO4 is a strong acid:
KOH(aq) + HClO4(aq) → KClO4(aq) + H2O(l)
Since KOH and HClO4 react in a 1:1 ratio, all the KOH will be neutralized by the HClO4 solution. This means that the number of moles of HClO4 added is the same as the number of moles of KOH initially present.
To solve this problem, we need information about the initial number of moles of KOH:
Molarity of KOH = 0.150 M
Volume of KOH = 30 mL = 0.030 L
Number of moles (n) of KOH = Molarity (M) × Volume (L)
n(KOH) = 0.150 mol/L × 0.030 L
n(KOH) = 0.0045 mol
Now we can calculate the pH after each volume of acid is added:
1) 30 mL of acid added (0 mL of HClO4 remaining):
The number of moles of HClO4 = 0.0045 mol (same as KOH)
The volume of solution = 30 mL + 30 mL = 60 mL = 0.060 L
Molarity of HClO4 = (moles of HClO4) / (volume of solution)
= 0.0045 mol / 0.060 L
= 0.075 M
Since HClO4 is a strong acid, it will dissociate completely, giving H+ ions in solution.
pH = -log[H+]
= -log[0.075]
= 1.12
2) 35 mL of acid added (5 mL of HClO4 remaining):
The number of moles of HClO4 = 0.0045 mol (same as KOH)
The volume of solution = 30 mL + 35 mL = 65 mL = 0.065 L
Molarity of HClO4 = (moles of HClO4) / (volume of solution)
= 0.0045 mol / 0.065 L
= 0.069 M
pH = -log[H+]
= -log[0.069]
= 1.16
3) 36 mL of acid added (6 mL of HClO4 remaining):
The number of moles of HClO4 = 0.0045 mol (same as KOH)
The volume of solution = 30 mL + 36 mL = 66 mL = 0.066 L
Molarity of HClO4 = (moles of HClO4) / (volume of solution)
= 0.0045 mol / 0.066 L
= 0.068 M
pH = -log[H+]
= -log[0.068]
= 1.16
4) 37 mL of acid added (7 mL of HClO4 remaining):
The number of moles of HClO4 = 0.0045 mol (same as KOH)
The volume of solution = 30 mL + 37 mL = 67 mL = 0.067 L
Molarity of HClO4 = (moles of HClO4) / (volume of solution)
= 0.0045 mol / 0.067 L
= 0.067 M
pH = -log[H+]
= -log[0.067]
= 1.17
5) 40 mL of acid added (10 mL of HClO4 remaining):
The number of moles of HClO4 = 0.0045 mol (same as KOH)
The volume of solution = 30 mL + 40 mL = 70 mL = 0.070 L
Molarity of HClO4 = (moles of HClO4) / (volume of solution)
= 0.0045 mol / 0.070 L
= 0.064 M
pH = -log[H+]
= -log[0.064]
= 1.19
So, the pH after adding the given volumes of acid are as follows:
1) pH = 1.12
2) pH = 1.16
3) pH = 1.16
4) pH = 1.17
5) pH = 1.19