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.
millimols KOH = 30 mL x 0.150 = 4.50
millimols HClO4 = mL x 0.125M = ??
..........HClO4 + KOH ==> KClO4 + H2O
begin..... 0......4.50........0.......0
add 30....3.75....0....................
react....-3.75...-3.75....+3.75...+3.75
final......0......0.75....+3.75...+3.75
HClO4 is a strong acid; KOH is a strong base. So the equivalence point will be 7.0 and neither cation nor anion will be hydrolyzed. Thus, the pH for the 30 mL addition will be determined by the excess KOH which will be 0.75 mmoles/(30+30 mL) = 0.75/60 =0.0125 = (OH^-), a pOH of 1.90, and a pH of 12.
Continue with the additions. It will be helpful if you determine where the equivalence point is.
To calculate the pH after different volumes of acid have been added, we need to determine the concentrations of both the KOH and HClO4 solutions as they react with each other.
Step 1: Determine the initial moles of KOH present:
Initial moles of KOH = initial volume of KOH (in L) x initial concentration of KOH (in M)
Initial moles of KOH = 0.030 L x 0.150 M
Initial moles of KOH = 0.0045 moles
Step 2: Determine the number of moles of HClO4 added:
Number of moles of HClO4 = volume of HClO4 added (in L) x concentration of HClO4 (in M)
For 30 mL of HClO4 added (equivalent to 0.030 L):
Number of moles of HClO4 = 0.030 L x 0.125 M
Number of moles of HClO4 = 0.00375 moles
Step 3: Determine the limiting reagent:
The limiting reagent is the one that is completely consumed in the reaction. In this case, KOH and HClO4 react in a 1:1 ratio, so the limiting reagent is the one with fewer moles. In this scenario, KOH has fewer moles (0.0045 moles).
Step 4: Determine the remaining moles of KOH after reacting with HClO4:
Remaining moles of KOH = initial moles of KOH - moles of HClO4 added
Remaining moles of KOH = 0.0045 moles - 0.00375 moles
Remaining moles of KOH = 0.00075 moles
Step 5: Determine the volume of the resulting solution:
Volume of the resulting solution = initial volume of KOH + volume of HClO4 added
Volume of the resulting solution = 0.030 L + 0.030 L
Volume of the resulting solution = 0.060 L
Step 6: Determine the concentration of the resulting solution:
Concentration of the resulting solution = remaining moles of KOH / volume of the resulting solution
Concentration of the resulting solution = 0.00075 moles / 0.060 L
Concentration of the resulting solution = 0.0125 M
Step 7: Calculate the pOH of the resulting solution:
pOH = -log10(concentration of OH- ions)
pOH = -log10(0.0125 M)
Step 8: Calculate the pH of the resulting solution:
pH = 14 - pOH
Now, let's calculate the pH after each volume of HClO4 is added:
For 30 mL (0.030 L) of HClO4 added:
pOH = -log10(0.0125 M)
pH = 14 - pOH
For 35 mL (0.035 L) of HClO4 added:
pOH = -log10(0.0125 M)
pH = 14 - pOH
For 36 mL (0.036 L) of HClO4 added:
pOH = -log10(0.0125 M)
pH = 14 - pOH
For 37 mL (0.037 L) of HClO4 added:
pOH = -log10(0.0125 M)
pH = 14 - pOH
For 40 mL (0.040 L) of HClO4 added:
pOH = -log10(0.0125 M)
pH = 14 - pOH
Please note that in each case, the pOH value is the same since the concentration of OH- ions in the resulting solution remains constant.
To calculate the pH at different volumes of acid added, we need to consider the reaction between KOH (base) and HClO4 (acid). The balanced equation is:
KOH + HClO4 -> KClO4 + H2O
Since KOH is a strong base and HClO4 is a strong acid, they react completely to form the products. This means that the number of moles of KOH initially present will be equal to the number of moles of KClO4 formed.
Let's break down the problem step by step:
1) Initial moles of KOH:
The given concentration of KOH is 0.150 M, and the volume of the solution is 30 mL. To calculate the initial moles of KOH, we can use the formula:
moles = concentration (M) * volume (L)
Therefore,
moles of KOH = 0.150 M * (30 mL / 1000 mL/L) = 0.0045 moles
2) Volume of HClO4 added:
For each volume of HClO4 added, we can calculate the total volume of the reaction solution. Note that the total volume increases as more acid is added.
3) Moles of HClO4:
To calculate the number of moles of HClO4, we can use the same formula as before:
moles = concentration (M) * volume (L)
4) Moles of KOH remaining:
Since KOH reacts with HClO4 in a 1:1 ratio, the remaining moles of KOH can be calculated by subtracting the moles of HClO4 used from the initial moles of KOH:
moles of KOH remaining = initial moles - moles of HClO4 used
5) Concentration of KOH remaining:
To calculate the concentration of KOH remaining, divide the moles of KOH remaining by the new total volume (original volume + volume of acid added):
concentration (M) = moles remaining / total volume (L)
6) Calculate pOH:
pOH is defined as -log10 of the hydroxide ion concentration. Since we are dealing with a base, pOH can be used to find the pH. The equation to convert pOH to pH is:
pH = 14 - pOH
Note: The pOH in this case is the negative logarithm of the hydroxide ion concentration derived from the remaining KOH concentration.
7) Calculate pH:
Using the pH equation, substitute the pOH value to obtain the pH.
By following these steps, we can calculate the pH after each specific volume of acid has been added.