Ka for hypochlorous acid,HCLO, IS 3.0*10^-8. Calculate the pH after 10.0,20.0,30.0, and 40.0mL OF 0.100M NaOH have been added to 40.0mL of 0.100M HCLO..
The pH of HClO at the beginning of the titration is calculated from HClO as a single weak acid.
Additions of NaOH form NaClO, which is a weak acid and its salt, therefore, the Henderson-Hasselbalch equation is used.
At the equivalence point, the acid is completely neutralized and the pH is determined by the hydrolysis of the salt.
After the equivalence point, the pH is determined by the excess amount of NaOH added.
To calculate the pH after adding NaOH to HCLO, we need to determine the concentrations of the components in the resulting solution at each step.
Given:
Initial volume of HCLO solution (V1) = 40.0 mL
Initial concentration of HCLO (C1) = 0.100 M
Volume of NaOH solution added (V2) = 10.0 mL, 20.0 mL, 30.0 mL, 40.0 mL
Concentration of NaOH solution(C2) = 0.100 M
Step 1: Determine the moles of HCLO initially present.
Moles of HCLO = C1 * (V1 / 1000) (converting mL to L)
Step 2: Determine the moles of NaOH added.
Moles of NaOH = C2 * (V2 / 1000) (converting mL to L)
Step 3: Determine the moles of HCLO remaining after reaction with NaOH.
Moles of HCLO remaining = Moles of HCLO initially present - Moles of NaOH added
Step 4: Determine the volume of the final solution.
Volume of final solution = V1 + V2 (summing the volumes)
Step 5: Determine the concentration of HCLO in the final solution.
Concentration of HCLO in the final solution = Moles of HCLO remaining / (Volume of final solution / 1000) (converting L to mL)
Step 6: Calculate the pH using the Ka expression for HCLO.
pH = -log10(sqrt(Ka * [HCLO]))
Now, let's calculate the pH at each step.
For 10.0 mL of NaOH added:
Step 1:
Moles of HCLO = 0.100 M * (40.0 mL / 1000) = 0.00400 moles
Step 2:
Moles of NaOH = 0.100 M * (10.0 mL / 1000) = 0.00100 moles
Step 3:
Moles of HCLO remaining = 0.00400 moles - 0.00100 moles = 0.00300 moles
Step 4:
Volume of final solution = 40.0 mL + 10.0 mL = 50.0 mL = 0.050 L
Step 5:
Concentration of HCLO in the final solution = 0.00300 moles / (0.050 L / 1000) = 0.0600 M
Step 6:
pH = -log10(sqrt((3.0*10^-8) * (0.0600 M))) = 4.08
Repeat the above steps for 20.0 mL, 30.0 mL, and 40.0 mL of NaOH added to find the pH at each step.
To calculate the pH after adding NaOH to HCLO, we need to consider the acid-base reaction that occurs between them. In this case, HCLO is a weak acid and NaOH is a strong base.
The balanced chemical equation for the reaction between HCLO and NaOH is:
HCLO + NaOH -> NaCLO + H2O
Since NaOH is a strong base, it completely dissociates into Na+ and OH- ions. Therefore, we can assume that the concentration of OH- ions is equal to the concentration of NaOH.
Initially, we have 40.0 mL of 0.100 M HCLO. The number of moles of HCLO can be calculated by multiplying the initial concentration by the volume:
moles HCLO = 0.100 M * 0.040 L = 0.004 moles
Since HCLO dissociates into H+ and CLO-, and the ratio between them is 1:1, we also have 0.004 moles of H+.
Now, let's calculate the pOH after adding 10.0 mL of 0.100 M NaOH:
moles NaOH = 0.100 M * 0.010 L = 0.001 moles
Since NaOH dissociates into Na+ and OH-, we have 0.001 moles of OH-. The final concentration of OH- ions can be calculated by dividing the moles by the total volume:
[OH-] = 0.001 moles / (40.0 mL + 10.0 mL) = 0.001 moles / 0.050 L = 0.020 M
The pOH can be calculated using the equation:
pOH = -log[OH-] = -log(0.020) = 1.70
Since pH + pOH = 14, we can calculate the pH:
pH = 14 - pOH = 14 - 1.70 = 12.30
We repeat the above calculations for the volumes of 20.0 mL, 30.0 mL, and 40.0 mL of 0.100 M NaOH added.
For 20.0 mL NaOH:
moles NaOH = 0.100 M * 0.020 L = 0.002 moles
[OH-] = 0.002 moles / (40.0 mL + 20.0 mL) = 0.002 moles / 0.060 L = 0.033 M
pOH = -log(0.033) = 1.48
pH = 14 - 1.48 ≈ 12.52
For 30.0 mL NaOH:
moles NaOH = 0.100 M * 0.030 L = 0.003 moles
[OH-] = 0.003 moles / (40.0 mL + 30.0 mL) = 0.003 moles / 0.070 L = 0.043 M
pOH = -log(0.043) = 1.37
pH = 14 - 1.37 ≈ 12.63
For 40.0 mL NaOH:
moles NaOH = 0.100 M * 0.040 L = 0.004 moles
[OH-] = 0.004 moles / (40.0 mL + 40.0 mL) = 0.004 moles / 0.080 L = 0.050 M
pOH = -log(0.050) = 1.30
pH = 14 - 1.30 = 12.70
Therefore, the pH after adding 10.0 mL, 20.0 mL, 30.0 mL, and 40.0 mL of 0.100 M NaOH are approximately 12.30, 12.52, 12.63, and 12.70, respectively.