Calculate the pH at each stage in the titration of 25.0 ml 0.10 M KOH(aq)
with 0.25 M HCl(aq).
1. Initiall ( before titration).
2. After addition of 5 ml of HCl.
3. After addition of 9.5 ml of HCl.
4. At the equivalent point.
5. After addition of 12 ml of HCl
1. Initial pH: 14 - pOH = 14- [-log(0.10 M KOH)]
To calculate the pH at each stage in the titration, you need to consider the reaction between KOH and HCl. The balanced equation for this reaction is:
KOH(aq) + HCl(aq) → KCl(aq) + H2O(l)
Here are the calculations for each stage:
1. Initial (before titration): Since you are starting with 25.0 ml of 0.10 M KOH, calculate the moles of KOH:
moles KOH = volume (L) × concentration (M)
= 0.025 L × 0.10 M
= 0.0025 moles
Since KOH is a strong base, it completely dissociates in water, giving the same number of hydroxide ions (OH-) as the number of moles of KOH.
Hence, the concentration of OH- ions is also 0.10 M.
The pOH can be calculated using the equation:
pOH = -log[OH-]
= -log(0.10)
= 1.0
Then, the pH can be calculated using the equation:
pH = 14 - pOH
= 14 - 1.0
= 13.0
So, the pH before titration is 13.0.
2. After addition of 5 ml of HCl: In this stage, you are adding 5 ml of 0.25 M HCl to the solution.
Since HCl is a strong acid, it completely dissociates in water, giving the same number of hydrogen ions (H+) as the number of moles of HCl.
moles HCl = volume (L) × concentration (M)
= 0.005 L × 0.25 M
= 0.00125 moles
The stoichiometry of the reaction between KOH and HCl is 1:1, meaning that for every 1 mole of HCl, 1 mole of OH- is consumed.
So, 0.00125 moles of OH- ions are neutralized.
Now, subtract the moles of OH- neutralized from the initial moles of OH-:
0.0025 moles - 0.00125 moles = 0.00125 moles of OH- remaining
Convert this to concentration:
concentration OH- = remaining moles OH- / total volume (L)
= 0.00125 moles / 0.030 L (25 ml + 5 ml)
= 0.0417 M
Now, calculate pOH and then pH as done in the initial stage.
pOH = -log[OH-]
= -log(0.0417)
= 1.38
pH = 14 - pOH
= 14 - 1.38
= 12.62
So, the pH after adding 5 ml of HCl is 12.62.
3. After addition of 9.5 ml of HCl: Follow the same procedure as in the previous stage, but now subtract 9.5 ml from the total volume (30 ml).
4. At the equivalent point: The equivalent point is reached when the moles of HCl added are equal to the moles of KOH initially present.
Calculate the moles of HCl based on the volume added up to this point, and use the stoichiometry to find the corresponding amount of OH- ions neutralized.
At the equivalent point, all the OH- ions have been neutralized and converted to water, so the concentration of OH- ions is zero.
Since pH = 14 - pOH, the pH at the equivalent point is 14.
5. After addition of 12 ml of HCl: Follow the same procedure as in the previous stages, but now subtract 12 ml from the total volume (30 ml).
Remember to calculate moles of HCl, moles of OH- neutralized, remaining moles of OH-, concentration of OH-, pOH, and pH.
I hope this explanation helps you understand how to calculate the pH at each stage in the titration.