1. What is the pH when enough 0.10 M Base (in mL) is added to neutralize 8 mL of 0.10 M Acid?
Complete the following table.
NaOH HC2H3O2
2. What is the pH of the solution created by combining 2.60 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
mL NaOH pH wHCl pH wHC2H3O2
2.60
Complete the table below:
What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
mL NaOH pH wHCl pH wHC2H3O2
2.60
3. What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in lab)?
mL NaOH pH wHCl pH wHC2H3O2
12.60
To determine the pH values in each scenario, we need to calculate the moles of acid and base present, and then use the stoichiometry of the reaction between the acid and base to determine the resulting concentration of the conjugate acid and base.
1. What is the pH when enough 0.10 M Base (in mL) is added to neutralize 8 mL of 0.10 M Acid?
To neutralize 8 mL of 0.10 M acid, we need an equal number of moles of 0.10 M base. Since the concentrations of the acid and base are the same, when neutralized, they will form equal concentrations of their conjugate acid and conjugate base.
Step 1: Calculate the moles of acid and base:
Moles of acid = Volume of acid (in L) × Concentration of acid
= 8 mL × (1 L / 1000 mL) × 0.10 M
= 0.0008 moles
Moles of base = Moles of acid (since they are stoichiometrically equivalent)
= 0.0008 moles
Step 2: Calculate the resulting concentration of the conjugate acid and base:
Since the acid and base are of the same concentration, when neutralized, they will form equal concentrations of the conjugate acid and conjugate base.
Concentration of the conjugate acid = Moles of acid / Total volume (in L)
= 0.0008 moles / (8 mL + Volume of base added) × (1 L / 1000 mL)
Step 3: Calculate the pH from the concentration of the acid:
pH = -log10(concentration of the conjugate acid)
Complete the table:
NaOH HC2H3O2
2. What is the pH of the solution created by combining 2.60 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HCl(aq)? with 8.00 mL of the 0.10 M HC2H3O2(aq)?
To determine the pH, we need to calculate the concentration of the resulting solution.
Step 1: Calculate the moles of acid and base:
Moles of acid = Volume of acid (in L) × Concentration of acid
= 8 mL × (1 L / 1000 mL) × 0.10 M
= 0.0008 moles
Moles of base = Volume of base (in L) × Concentration of base
= 2.60 mL × (1 L / 1000 mL) × 0.10 M
= 0.00026 moles
Step 2: Calculate the total volume of the resulting solution:
Total volume = Volume of acid + Volume of base
= 8 mL + 2.60 mL
= 10.60 mL
Step 3: Calculate the resulting concentration of the resulting solution:
Concentration of the resulting solution = (Moles of acid + Moles of base) / Total volume
= (0.0008 moles + 0.00026 moles) / (10.60 mL) × (1 L / 1000 mL)
Step 4: Calculate the pH from the concentration of the resulting solution:
pH = -log10(concentration of the resulting solution)
Complete the table:
mL NaOH pH wHCl pH wHC2H3O2
2.60
3. What are the pH values if you take into account that the 8.00 mL of 0.10 M Acid was first diluted with 100 mL of water (like it will be in the experiment you perform in the lab)?
To account for the dilution of the acid with water, we need to recalculate the concentration of the acid before proceeding with the pH calculations.
Step 1: Calculate the new concentration of the acid after dilution:
Concentration of the acid after dilution = (Volume of undiluted acid / Total volume after dilution) × Concentration of undiluted acid
= (8 mL / (8 mL + 100 mL)) × 0.10 M
Step 2: Repeat the calculations from question 2 with the new concentration of the acid, keeping the rest of the values the same.
Complete the table:
mL NaOH pH wHCl pH wHC2H3O2
12.60
By following these steps, you should be able to determine the pH values for each scenario. Remember to substitute the appropriate values and units into the equations.