44.70 ml of 0.100 M NaOH are required to completely neutralize 50.00 ml of a weak monoprotic acid, HA. When 34.43 ml of NaOH are added to another 50.00 ml of the acid, the pH reading was 4.80. Calculate Ka for the acid using the partial neutralization method.
anyone else get the same answer?
0.0447 L total neut'n with NaOH
0.03443 L partial neut'n
0.1 M NaOH
0.00447 moles T Total
0.04720169 T (M)
0.003443 moles P Partial
0.040779344 P (M)
4.8 pH
1.58489E-05 [H] = [A-]
3.91117E-08 Ka
Please ignore my previous answer in which I assumed that [A-] = [H+]. I think [A-] should be calculated using the partial titration data, and [H+] is known from the pH. My latest attempt is below:
0.0447 L total neut'n with NaOH
0.03443 L partial neut'n
0.1 M NaOH
0.00447 moles T Total
0.04720169 T (M)
0.003443 moles P Partial
0.040779344 P (M)
4.8 pH
1.58489E-05 [H]
0.040779344 [A-] = P
0.000100634 Ka
To calculate Ka for the acid using the partial neutralization method, we need to use the Henderson-Hasselbalch equation and the information provided.
Here's the step-by-step process to solve this problem:
Step 1: Calculate the moles of NaOH used in the neutralization reaction.
Given:
Volume of NaOH = 44.70 ml
Concentration of NaOH = 0.100 M
Use the formula: moles = concentration × volume
moles of NaOH = 0.100 M × 0.04470 L
moles of NaOH = 0.00447 moles
Step 2: Calculate the moles of acid in the 50.00 ml sample.
Given:
Volume of acid = 50.00 ml
We assume that the acid is monoprotic, so it reacts with NaOH in a 1:1 ratio. Therefore, the moles of acid will be the same as the moles of NaOH used in the neutralization reaction: 0.00447 moles.
Step 3: Calculate the initial concentration of the acid.
Given:
Volume of acid = 50.00 ml
Moles of acid = 0.00447 moles
Use the formula: concentration = moles/volume
Initial concentration of the acid = 0.00447 moles / 0.0500 L
Initial concentration of the acid = 0.0894 M
Step 4: Calculate the concentration of the acid after the partial neutralization.
Given:
Volume of acid = 50.00 ml
Volume of NaOH added = 34.43 ml
Subtract the volume of NaOH added from the total volume of the acid to get the volume of remaining acid:
Volume of remaining acid = 50.00 ml - 34.43 ml
Volume of remaining acid = 15.57 ml
Step 5: Calculate the moles of remaining acid.
Given:
Volume of remaining acid = 15.57 ml
Moles of acid before partial neutralization = 0.00447 moles
Use the formula: moles = concentration × volume
Moles of acid after partial neutralization = concentration × volume
Moles of acid after partial neutralization = 0.0894 M × 0.01557 L
Moles of acid after partial neutralization = 0.00139 moles
Step 6: Calculate the concentration of the acid after the partial neutralization.
Given:
Volume of remaining acid = 15.57 ml
Moles of acid after partial neutralization = 0.00139 moles
Use the formula: concentration = moles/volume
Concentration of the acid after partial neutralization = 0.00139 moles / 0.01557 L
Concentration of the acid after partial neutralization = 0.0894 M
Step 7: Calculate the pOH of the solution.
Given:
pH reading = 4.80
Use the formula: pOH = 14 - pH
pOH = 14 - 4.80
pOH = 9.20
Step 8: Calculate the concentration of hydroxide ions (OH-) in the solution.
Given:
pOH = 9.20
Use the formula: concentration of OH- = 10^(-pOH)
Concentration of OH- = 10^(-9.20)
Concentration of OH- = 6.31 × 10^(-10) M
Step 9: Calculate the concentration of acid (HA-) in the solution after the partial neutralization.
Given:
Concentration of OH- = 6.31 × 10^(-10) M
Concentration of the acid after partial neutralization = 0.0894 M
Use the formula: Ka = (concentration of HA-) × (concentration of OH-) / (concentration of A-)
Since we assume the acid is weak, we can neglect the concentration of A- (conjugate base) compared to HA-.
Therefore, Ka ≈ (concentration of HA-) × (concentration of OH-)
Ka ≈ (0.0894 M) × (6.31 × 10^(-10) M)
Step 10: Calculate Ka.
Ka ≈ 5.64 × 10^(-11) M
Therefore, the Ka value for the acid, HA, is approximately 5.64 × 10^(-11) M.