Consider the titration of 50.0 mL of 0.200 M HOCl by 0.200 M KOH for the next five questions (Ka for HOCl = 3.5 x 10-8).

Calculate the pH after 150.0 mL of KOH has been added.

KOH + HOCl => H2O + KCl

On these problems it helps to know where the equivalence point is located.
mL HOCl x M HOCl = mL KOH x M KOH
Substitute and solve for mL KOH. I get 50 mL NaOH needed. Therefore, you know this 150 is well past the equivalence point so the pH is determined by the excess KOH present
You started with 50 x 0.2 = 10 mmoles HOCl
You added 150 mL x 0.2 = 30 mmols KOH.

..........HOCl + KOH ==> KCl + H2O
I..........10.....0.......0.....0
add .............30...............
C.........-10...-10......+10....+10
E..........0.....20......+10....+10

Therefore, you have 20 mmols KOH in 150 + 50 = 200 mL solution.
(OH^-) = 20/200 = ?
pOH = -log(OH^-)
Convert pOH to pH.

Why did the acid go to the party?

Because it wanted to have a pH-enomenal time!

To calculate the pH after 150.0 mL of KOH has been added, we need to determine the moles of HOCl and KOH present in the solution.

First, let's calculate the moles of HOCl initially present:
Moles of HOCl = Molarity * Volume
= 0.200 M * 0.0500 L
= 0.0100 moles

Next, let's calculate the moles of KOH added:
Moles of KOH = Molarity * Volume
= 0.200 M * 0.1500 L
= 0.0300 moles

Since HOCl and KOH have a 1:1 stoichiometry, we can see that the moles of HOCl reacted with the moles of KOH.

Moles of HOCl remaining = Moles of HOCl initially - Moles of KOH added
= 0.0100 moles - 0.0300 moles
= -0.0200 moles (negative because KOH exceeds HOCl)

Now, let's calculate the concentration of HOCl after 150.0 mL of KOH has been added:
Concentration of HOCl = Moles of HOCl remaining / Volume
= -0.0200 moles / (0.0500 L + 0.1500 L)
= -0.0200 moles / 0.2000 L
= -0.100 M (negative because KOH exceeds HOCl)

Since the concentration cannot be negative, we know that all the HOCl has reacted with KOH. Therefore, the remaining solution contains only KOH.

To calculate the pH, we need to determine the concentration of OH- ions in the solution. Since KOH is a strong base, it completely dissociates into KO+ and OH- ions.

Concentration of OH- = Molarity of KOH * Volume of KOH added
= 0.200 M * 0.1500 L
= 0.0300 M

Since we have the concentration of OH- ions, we can calculate the pOH using the formula:

pOH = -log10[OH-]

pOH = -log10[0.0300]
= 1.52

Finally, we can calculate the pH using the equation:

pH + pOH = 14

pH + 1.52 = 14
pH = 14 - 1.52
pH = 12.48

Therefore, the pH after 150.0 mL of KOH has been added is approximately 12.48.

To calculate the pH after 150.0 mL of KOH has been added, we need to consider the reaction between HOCl and KOH.

The balanced equation for this reaction is:

HOCl + KOH -> KCl + H2O

From the balanced equation, we can see that 1 mole of HOCl reacts with 1 mole of KOH to form 1 mole of KCl and 1 mole of water.

Let's calculate the number of moles of HOCl in the initial 50.0 mL solution.

moles of HOCl = volume of solution (in L) x molarity of HOCl
= 0.050 L x 0.200 M
= 0.010 mol

Since we have equal molar amounts of HOCl and KOH, the reaction will consume 0.010 mol of KOH when 0.010 mol of HOCl is consumed.

Now, let's calculate the number of moles of KOH in 150.0 mL of the KOH solution that has been added.

moles of KOH = volume of solution (in L) x molarity of KOH
= 0.150 L x 0.200 M
= 0.030 mol

As we mentioned earlier, 0.010 mol of KOH will react with the 0.010 mol of HOCl to form KCl and water. Therefore, the remaining 0.020 mol of KOH will be in excess.

To find the resulting concentration of KOH, we need to consider the total volume of the solution after adding the 150.0 mL of KOH.

total volume of solution = initial volume of HOCl + volume of KOH added
= 0.050 L + 0.150 L
= 0.200 L

Concentration of the excess KOH = moles of excess KOH / total volume of solution
= 0.020 mol / 0.200 L
= 0.100 M

Now, let's calculate the concentration of the remaining HOCl by subtracting the consumed moles of HOCl from the initial moles of HOCl.

moles of remaining HOCl = initial moles of HOCl - consumed moles of HOCl
= 0.010 mol - 0.010 mol
= 0 mol

Since all the HOCl has been consumed, the concentration of HOCl is now zero.

The key to calculating the pH after 150.0 mL of KOH has been added lies in the fact that this reaction produces KCl and H2O. KCl is a salt and will not affect the pH. Therefore, we only need to consider the reaction of water with HOCl.

HOCl + H2O -> H3O+ + OCl-

This is an acid-base reaction, where HOCl acts as an acid, donating a proton (H+) to water to form hydronium ions (H3O+), and leaving behind hypochlorite ions (OCl-).

Given that the Ka for HOCl is 3.5 x 10^-8, we can determine the concentration of H3O+ using the equation for Ka:

Ka = [H3O+][OCl-] / [HOCl]

Since we have already determined that the concentration of HOCl is zero, we can omit it from the equation:

Ka = [H3O+][OCl-] / 0

This means that the concentration of H3O+ is zero, and therefore the pH is infinity.

So, the pH after 150.0 mL of KOH has been added is infinity.