Question

A) 25.00 mL of 0.100 M concentrated HCL titrated with 0.100 M NaOH solution. i) Write the chemical reaction for this titration. ii) Find the initial pH of the solution. iii) Find the pH, a) Before equivalence point (after b) At the equivalence point c) After the equivalence point (after adding 26.00mL of titrant) iv) Draw the titration curve. B) What volume of 0.180 M solution of KOH is needed to titrate 30.0 mL of 0.200 M H2SO4? (i.e. consider the stoichiometry)​

Answer 1

A)

i) The chemical reaction for the titration of HCl with NaOH is:

HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

ii) The initial pH of the solution can be calculated using the following formula:

pH = -log[H+]

Since HCl is a strong acid and is completely dissociated in water, the concentration of H+ ions in the solution is equal to the concentration of HCl. Therefore, the initial pH of the solution is:

pH = -log(0.100) = 1.00

iii)

a) Before equivalence point (after adding 0.00 mL of titrant):

Before the addition of any NaOH solution, the pH of the HCl solution is the same as the initial pH calculated in (ii), which is 1.00.

b) At the equivalence point (after adding 25.00 mL of titrant):

At the equivalence point, the moles of NaOH added is equal to the moles of HCl initially present. Therefore, the number of moles of NaOH is:

moles of NaOH = concentration of NaOH x volume of NaOH

moles of NaOH = 0.100 x 0.025 = 0.0025 mol

Since HCl and NaOH react in a 1:1 ratio, the number of moles of HCl initially present is also 0.0025 mol. The total volume of the solution after the addition of 25.00 mL of NaOH is:

total volume = volume of HCl + volume of NaOH

total volume = 0.025 + 0.025 = 0.050 L

Therefore, the concentration of HCl at the equivalence point is:

concentration of HCl = moles of HCl / total volume

concentration of HCl = 0.0025 / 0.050

concentration of HCl = 0.050 M

Since HCl is a strong acid, it is completely dissociated in water. Therefore, the concentration of H+ ions at the equivalence point is 0.050 M. Using the formula for pH, the pH at the equivalence point is:

pH = -log(0.050) = 1.30

c) After the equivalence point (after adding 26.00 mL of titrant):

After the equivalence point, there is excess NaOH in the solution. The excess NaOH will react with water to produce OH- ions. The concentration of OH- ions can be calculated using the following formula:

moles of excess NaOH = concentration of NaOH x volume of NaOH added - moles of HCl initially present

moles of excess NaOH = 0.100 x (0.026 - 0.025) - 0.0025

moles of excess NaOH = 0.00075 mol

Since NaOH and water react in a 1:1 ratio to produce OH- ions, the concentration of OH- ions in the solution is:

concentration of OH- = moles of excess NaOH / total volume

concentration of OH- = 0.00075 / 0.051

concentration of OH- = 0.0147 M

Using the formula for pH, the pH after the equivalence point is:

pH = 14 - (-log[OH-]) = 11.84

Step-by-step explanation: