Question

Answer the following sections from the following data of the pH constant for the sub-halogen acids at 0.1M:

Ka (HClO) = 3.5x10 ^ -8
Ka (HBrO) = 2.0x10 ^ -9

a. Calculate the pH level of 1.00 liters of water dissolved in 2.5 g of NaOCl.
b. Do you expect that the pH level of 1.00 liters of water that has been dissolved in 2.5 grams NaOBr, will be higher, lower or equal to the result of the section a?
c. Calculate the pH level at the stoichiometric point after adding 29.80 mL NaOH solution at 0.0567M concentration to 25.00 ml solution of subcellular acid.
d. Do you expect a higher or lower pH level at the stochometric point of sub-bromitic acid under the same titration conditions of section 3?

Answer 1

a) The dissolved moles of salt are calculated:

n Salt = g / MM = 2.5 / 74.44 = 0.034 mol

The ClO- ion concentration is 0.034 M, hydrolysis occurs:

ClO- + H2O = HClO + OH-

You have Kb:

Kb = Kw / Ka = 10 ^ -14 / 3.5x10 ^ -8 = 2.9x10 ^ -7

It has its expression:

Kb = [HClO] * [OH-] / [ClO-] = X ^ 2 / 0.034 - X

It is assumed that - X is negligible and clears:

X = [OH-] = √Kb * 0.034 = √2.9x10 ^ -7 * 0.034 = 9.93x10 ^ -5 M

POH and pH are calculated:

pOH = - log 9.93x10 ^ -5 = 4

pH = 14 - 4 = 10

b) HBrO is less acidic, since it has a lower Ka, it is assumed that it will have a higher pH than case A.

c) The moles of NaOH added are calculated:

n NaOH = M * V = 0.0567 * 0.0298 = 1.7x10 ^ -3 mol

The initial moles of acid are calculated:

n Acid = 0.1 * 0.025 = 0.0025 mol

The NaOH reacts with the acid, forming salt, the remaining moles of acid are calculated:

n Acid = 0.0025 - 0.0017 = 0.0008 mol

By means of the henderson hasselbach equation, the solution's pH is calculated:

pH = - log Ka + log (n Salt / n Acid) = - log (3.5x10 ^ -8) + log (0.0017 / 0.0008) = 7.78

d) The HBrO is less acidic, therefore the pH will be higher in its case.

If you liked the answer, please rate it in a positive way, you would help me a lot, thank you.