Question

22. Considering the titration of a 25.0 mL sample of 0.115 NaOH with 0.100 M HCl,

calculate,
i) the initial pH (before adding any HCl solution).
ii the volume of acid required to reach the equivalence point.
iii) the pH after addition of 5.0 mL HCl.
iv) the pH at the equivalence point.

Answer 1

The initial pH can be calculated using the concentration and volume of NaOH. The pH after the addition of 5.0 mL HCl can also be calculated using the concentration of hydrogen ions. The volume of acid required to reach the equivalence point and the pH at the equivalence point can be determined through equations and the titration curve.

Step-by-step explanation:

The initial pH of the solution before adding any HCl can be calculated using the equation:

pH = -log[OH-]

Since the concentration of NaOH is 0.115 M and the volume is 25.0 mL, the number of moles of NaOH can be calculated using the equation:

moles = concentration x volume

Therefore, the initial pH can be calculated by using the formula for pH and the concentration of hydroxide ions:

pH = -log(0.115/0.025)

After adding 5.0 mL of HCl to the solution, the volume of the solution has increased to 30.0 mL. To calculate the concentration of hydrochloric acid in the solution, we need to use the equation:

moles = concentration x volume

By substituting the values into the equation, we can find the concentration of hydrochloric acid in the solution:

concentration = moles/volume

Finally, the pH after the addition of 5.0 mL of HCl can be calculated using the formula for pH and the concentration of hydrogen ions:

pH = -log(concentration)

The volume of acid required to reach the equivalence point can be determined by plotting the titration curve for the reaction. The equivalence point is the point at which the number of moles of acid is equal to the number of moles of base in the solution. By using the equation:

volume of acid = concentration of acid x volume of acid

The pH at the equivalence point can be determined by using the equation:

pH = 7

Answer 2

i) The initial pH before adding any HCl solution is 13.06

ii) The volume of acid required to reach equivalence point is 28.75 mL

iii) The pH after addition of 5.0 mL HCl is 12.899

iv) The pH at the equivalence point is 7

Step-by-step explanation:

i) Here we have the concentration of the sample = 0.115M NaOH

Therefore pH of NaOH

pOH = -log[OH⁻] = -log[0.115] = 0.9393

pH + pOH = 14

∴ pH = 14 - pOH = 14 - 0.9393 = 13.06

The initial pH before adding any HCl solution = 13.06

ii) The equivalence point is the point of complete neutralization, therefore;

NaOH + HCl → NaCl + H₂O

One mole of NaOH reacts with one mole of HCl to form one mole of NaCl and one mole of H₂O

Therefore, 0.115 Mole of NaOH will be completely neutralized by 0.115 mole of HCl, hence we have

25 × 0.115 moles of NaOH ≡ x × 0.1 of HCl

Where:

x = Volume of HCl required in mL

∴ x = 25×0.115/0.1 = 28.75 mL of HCl is required to reach equivalence point

The volume of acid required to reach equivalence point = 28.75 mL

iii) The pH after addition of 5.0 mL HCl is found as follows

The volume, y of NaOH that is completely neutralized by 5.0 mL HCl is given by the following expression

y×0.115 = 5×0.1

y = 5×0.1/0.115 = 4.35 mL

Therefore, since the total volume is now 25 + 5 = 30 mL and the amount of 0.115 M NaOH in the solution is 25 - 4.35 = 20.65 mL, the molarity of the solution is therefore;

20.65 mL/1000 × 0.115 = 0.002375 moles in 30 mL

Number of moles in one liter = 0.002375/(30/1000) = 0.0792 moles

Hence the new molarity of NaOH = 0.0792 M

The pH = 14 - pOH = 14 - (-log[OH])

= 14 - (-log(0.0792)) = 12.899

The pH after addition of 5.0 mL HCl = 12.899

iv) The pH at equivalence point after complete neutralization is the pH of a neutral solution hence

pH at the equivalence point = 7.