A 25.0 mL sample of 0.150 M hypochlorous acid is titrated with a 0.150 M NaOH solution. What is the pH at the equivalence point? The Kaof hypochlorous acid is 3.0x10^-8. a) 10.20
b) 7.00
c) 6.48
d) 7.52
e) 14.52

pH = 10.20

Explanation:

The HClO reacts with NaOH as follows:

HClO + NaOH → H2O + NaClO

Where HClO and NaOH react in a 1:1 reaction.

As the concentration of both reactions is the same and the reaction is 1:1, to reach equivalence point are required the same 25.0mL.

And the NaClO produced decreases its concentration in 2 because the volume is doubled.

The concentration of NaClO is: 0.150M / 2 = 0.075M

The equilibrium of NaClO is:

NaClO(aq) + H2O(l) ⇄ HClO(aq) + OH-(aq)

Where Kb of reaction is 1.0x10⁻¹⁴ / Ka =

1.0x10⁻¹⁴ / 3.0x10⁻⁸ = 3.33x10⁻⁷ = [HClO] [OH-] / [NaClO]

[NaClO] = 0.075M

As both HClO and OH- comes from the same equilibrium,

[HClO] = [OH-] = X

Where X is the reactoin coordinate

Replacing:

3.33x10⁻⁷ = [X] [X] / [0.075M]

2.5x10⁻⁸ = X²

X = 1.58x10⁻⁴M = [OH-]

pOH = -log [OH-]

pOH = 3.80

pH = 14 - pOH

B.  2 moles of water

Explanation:

The reaction expression is given as shown below:

            2H₂   +    O₂   →    2H₂O  

Number of moles of Hydrogen gas  = 2 moles

Number of moles of oxygen gas  = 2.5moles

Unknown:

Number of moles of water produced  = ?

Solution:

To solve this problem, we need to find the limiting reactant because this reactant will determine the extent of the reaction.

   From the balanced reaction expression:

   2 mole of hydrogen gas will need to combine with 1 mole of oxygen gas

   But we have been given 2.5 moles of oxygen gas

Oxygen gas is in excess and hydrogen gas is the limiting reactant.

So;

       2 moles of hydrogen gas will produce 2 moles of water