Calculate the pH for each of the following cases in the titration of 50.0 mL of 0.230 M HClO (aq) with 0.230 M KOH (aq). The ionization constant (Ka) for HClO is 4.00 x 10^-8.
(b) after addition of 25.0 mL of KOH
(d) after addition of 50.0 mL of KOH
To calculate the pH at each stage, we need to determine the moles of acid and base that have reacted.
First, let's calculate the moles of HClO (acid) initially present:
moles HClO = concentration of HClO x volume of HClO
= 0.230 M x 50.0 mL
= 0.0115 moles
Now, let's determine the moles of KOH (base) that have reacted in each case.
(b) After addition of 25.0 mL of KOH:
moles KOH = concentration of KOH x volume of KOH added
= 0.230 M x 25.0 mL
= 0.00575 moles
Since HClO and KOH react in a 1:1 ratio, the moles of HClO remaining will be the initial moles minus the moles of KOH reacted:
moles HClO remaining = initial moles HClO - moles KOH reacted
= 0.0115 moles - 0.00575 moles
= 0.00575 moles
To find the concentration of HClO remaining:
concentration HClO remaining = moles HClO remaining / volume of HClO remaining
= 0.00575 moles / (50.0 mL + 25.0 mL)
= 0.00575 moles / 0.075 L
= 0.0767 M
Now, calculate the pOH of the solution:
pOH = -log10 (OH- concentration)
pOH = -log10 (0.230 M) = 0.638
Finally, calculate the pH using the pOH:
pH = 14 - pOH = 14 - 0.638 = 13.362
Therefore, the pH after the addition of 25.0 mL of KOH is 13.362.
(d) After addition of 50.0 mL of KOH:
Using a similar approach, we find:
moles KOH = 0.230 M x 50.0 mL = 0.0115 moles
moles HClO remaining = 0.0115 moles - 0.0115 moles = 0 moles
concentration HClO remaining = 0 moles / (50.0 mL + 50.0 mL) = 0 moles / 0.1 L = 0 M
pOH = -log10 (OH- concentration) = -log10 (0.230 M) = 0.638
pH = 14 - pOH = 14 - 0.638 = 13.362
Therefore, the pH after the addition of 50.0 mL of KOH is 13.362.
To calculate the pH for each case in the titration, we need to consider the chemical reaction that occurs between HClO and KOH, and then determine the amount of excess or limiting reagent. From there, we can use the appropriate formulas to calculate the concentration of each species present and the resulting pH.
Let's start by writing the balanced chemical equation for the reaction between HClO and KOH:
HClO (aq) + KOH (aq) -> KClO (aq) + H2O (l)
(a) Before any KOH is added:
In this case, there is no KOH added yet, so the reaction has not started. Hence, the pH can be calculated by assuming 50.0 mL of 0.230 M HClO is present. To calculate the pH, we will use the equation for the ionization constant (Ka) of HClO:
Ka = [H+][ClO-] / [HClO]
Given that Ka = 4.00 x 10^-8 and the initial concentration of HClO is 0.230 M, the concentration of ClO- at the start is negligible compared to HClO. Therefore, we can assume that [HClO] = [H+]. Hence, we can rewrite the equation as:
Ka = [H+]^2 / [HClO]
Now solve for [H+]:
[H+]^2 = Ka x [HClO]
[H+]^2 = (4.00 x 10^-8) x (0.230)
Taking the square root of both sides:
[H+] = √ [(4.00 x 10^-8) x (0.230)]
Calculate the value of [H+].
(b) After addition of 25.0 mL of KOH:
In this case, we need to determine the excess reagent, if any, and calculate the concentrations accordingly. Since the initial concentration of both HClO and KOH is 0.230 M, and we have added half the volume of KOH, the reaction will be halfway through. Thus, we can assume that there is an equal stoichiometric amount of HClO and KOH, resulting in a neutral solution. Therefore, the pH is equal to 7.
(d) After addition of 50.0 mL of KOH:
With the addition of the full volume of KOH, we will have used all the HClO and have an excess of KOH. To determine the concentration of OH- ions, we will first calculate the initial moles of HClO and KOH:
moles of HClO = initial concentration of HClO x volume of HClO
moles of KOH = initial concentration of KOH x volume of KOH
Next, we will compare the moles of HClO and KOH to determine which one is limiting:
If moles of HClO < moles of KOH, HClO is limiting (excess KOH).
If moles of HClO > moles of KOH, KOH is limiting (excess HClO).
Since the initial concentrations and volumes are the same, the moles of HClO and KOH will also be the same, resulting in a neutral solution with a pH equal to 7.
To summarize:
- (b) After addition of 25.0 mL of KOH: pH = 7
- (d) After addition of 50.0 mL of KOH: pH = 7
You must learn to recognize where you are on the titration curve.
1. Start by calculating mL KOH to reach the equivalence point.
50.0 x 0.230 = mL x 0.230
Obviously it will be 50.0 mL KOH and that will answer part. That is determined by the hydrolysis of the salt, KClO.
.........ClO^- + HOH ==> HClO + OH^-
initial..0.150M............0.....0
change....-x...............x.....x
equil....0-.150-x...........x.....x
Kb for ClO^- = (Kw/Ka for HClO) = (x)(x)/(0.150-x)
Solve for x = (OH^-) and convert to pH.
For part a.
........HClO + KOH ==> KClO + H2O
....50*0.230 + 25.0*0.230 .....
mmol....11.5.. 5.75......0.....0
change..-5.75..-5.75....5.75...5.75
equil..4.25......0.......5.75..5.75
Ka = (H^+)(ClO^-)/(HClO)
Plug in x for (H^+), get HClO and ClO^- from the ICE chart, solve for x and convert to pH.
Alternatively, you can use the Henderson-Hasselbalch equation.