Calculate the pH for each of the following points in the titration of 50.0 mL of a 1.8 M H3PO3(aq) with 1.8 M KOH(aq).
before KOH
after addition of 25, 50,75,100 ml KOH
your explanation was extremely poor.
"I'm sure you've done hundreds of these"
If we did, then we wouldn't be looking up how to do the problem.
To calculate the pH at each point in the titration, we need to determine the number of moles of both the acid (H3PO3) and the base (KOH) at each point.
Before adding KOH:
At the start, 50.0 mL of a 1.8 M H3PO3 solution is present. To calculate the number of moles of H3PO3, we use the formula:
moles = concentration × volume
moles = 1.8 M × 0.050 L
moles = 0.09 mol
Since H3PO3 is a weak diprotic acid, it ionizes in two steps:
H3PO3 + H2O ⇌ H2PO3- + H3O+
H2PO3- + H2O ⇌ HPO32- + H3O+
Since H3PO3 is a weak acid, we can assume it barely dissociates, and hence, neglect the contribution of H3PO3 to the [H3O+].
After the addition of 25 mL KOH:
At this point, 25 mL of 1.8 M KOH solution is added. This will react with the acid to form water. The mole ratio between H3PO3 and KOH is 1:1. Therefore, the number of moles of KOH is:
moles = concentration × volume
moles = 1.8 M × 0.025 L
moles = 0.045 mol
Since KOH is a strong base, it completely dissociates in water:
KOH → K+ + OH-
The OH- will react with the H3O+ ion from the first dissociation of H3PO3.
H3O+ + OH- → 2H2O
This reaction consumes H3O+, resulting in an increase in the pH. However, at this point, the acid is still in excess, and the pH will be determined by the concentration of the acid.
After the addition of 50 mL KOH:
Now a total volume of 75 mL (50 mL H3PO3 + 25 mL KOH) is present. The number of moles of KOH added remains the same: 0.045 mol.
The mole ratio between H3PO3 and KOH is still 1:1, so the total moles of KOH is 0.045 mol.
Since H3PO3 is a diprotic acid, both moles of KOH will react with moles of the first ionization of H3PO3. Therefore, the number of moles of H3PO3 remaining is 0.09 mol - 0.045 mol = 0.045 mol.
After the addition of 75 mL KOH:
Now a total volume of 125 mL (50 mL H3PO3 + 75 mL KOH) is present.
The number of moles of KOH added remains the same: 0.045 mol.
Again, both moles of KOH will react with moles of the first ionization of H3PO3. Therefore, the number of moles of H3PO3 remaining is 0.09 mol - 0.045 mol - 0.045 mol = 0 mol.
After the addition of 100 mL KOH:
Now a total volume of 150 mL (50 mL H3PO3 + 100 mL KOH) is present.
The number of moles of KOH added remains the same: 0.045 mol.
Once again, both moles of KOH will react with moles of the first ionization of H3PO3. Therefore, the number of moles of H3PO3 remaining is 0.
Since there is no H3PO3 remaining, the pH will be determined solely by the concentration of OH-. The concentration of OH- can be determined by calculating the moles of OH- and dividing by the total volume of the solution:
Moles of OH- = 0.045 mol
Total volume = 0.150 L
Concentration of OH- = moles of OH- / total volume
Concentration of OH- = 0.045 mol / 0.150 L
Concentration of OH- = 0.30 M
Now we can determine the pOH using the equation:
pOH = -log10[OH-]
pOH = -log10(0.30)
pOH = 0.522
To calculate the pH, we use the equation:
pH + pOH = 14
pH + 0.522 = 14
pH = 14 - 0.522
pH = 13.478
Therefore, the pH after the addition of 100 mL KOH is approximately 13.478.
To calculate the pH at each point in the titration, we need to know the stoichiometry of the reaction between H3PO3 and KOH. H3PO3 is a diprotic acid, meaning it can donate two protons (H+ ions) in a reaction.
The reaction between H3PO3 and KOH can be represented as follows:
H3PO3(aq) + KOH(aq) → KH2PO3(aq) + H2O(l)
At the beginning of the titration, before any KOH is added, we have only H3PO3 in solution. The pH at this point can be calculated using the concentration of H3PO3.
To calculate the pH at each point after the addition of KOH, we need to determine the amount of H3PO3 that has reacted with KOH. This can be done by calculating the moles of KOH used in each step and using the stoichiometry of the reaction.
Given that the initial volume of H3PO3 is 50.0 mL and the concentration is 1.8 M, we can calculate the initial moles of H3PO3 as follows:
moles H3PO3 = volume (L) × concentration (M)
moles H3PO3 = 0.050 L × 1.8 M = 0.09 mol
Now, we can calculate the moles of KOH used at each point after the addition of 25 mL, 50 mL, 75 mL, and 100 mL of KOH.
For the first point, after adding 25 mL of KOH, we have:
moles KOH = volume (L) × concentration (M)
moles KOH = 0.025 L × 1.8 M = 0.045 mol
Since the stoichiometry of the reaction is 1:1 between H3PO3 and KOH, we can see that 0.045 mol of H3PO3 has reacted with KOH. This means that 0.045 mol of H3PO3 has been converted to KH2PO3. The remaining amount of H3PO3 can be calculated by subtracting the moles of KOH reacted from the initial moles of H3PO3:
moles H3PO3 remaining = initial moles of H3PO3 - moles of KOH reacted
moles H3PO3 remaining = 0.09 mol - 0.045 mol = 0.045 mol
Now, we can calculate the concentration of H3PO3 at this point:
concentration H3PO3 = moles of H3PO3 remaining / volume (L)
concentration H3PO3 = 0.045 mol / 0.050 L = 0.9 M
Using this concentration of H3PO3, we can use the equation for a diprotic acid to calculate the pH. However, it's important to note that H3PO3 is a weak acid, so we need to consider the dissociation of the acid.
H3PO3 ⇌ H+ + H2PO3-
To calculate the pH, we need to know the equilibrium constant (Ka) for the dissociation of H3PO3. Unfortunately, the Ka value for H3PO3 is not readily available in the question. Therefore, we cannot accurately determine the pH for each point in the titration without the Ka value.
I hope this explanation helps in understanding how to calculate the pH in a titration, but please note that the specific values required for an accurate calculation are missing in this case.
The secret to these is to know where you are on the titration curve.
First, find the equivalence points. There are three of them.
Write the equation and balance it.
Mols H3PO4 = M x L = ?
mols KOH = mols H3PO4 that (look at the coefficients in the equation).
mols KOH = M KOH/L KOH. You know mols and M, solve for L.
Then at zero mL KOH you have 1.8M H3PO4. I'm sure you've done hundred o these.
Between zero and the first equivalence point use th Henderson-Hasselbalch equation.
At each eq pt use the hydrolysis of the salt. For the first one it will be
.........H2PO4^- + HOH =-=> H3PO3 + H2O
I.........0.09...............0........0
C.........-x.................x........x
E.........0.09-x..............x........x
Kb for H2PO4^- = (Kw/k1 for H3PO4) = (x)(x)/(0.09-x) and solve for x = OH^-. Convert to pH.
The 0.09 comes from this.
You had 50 mL x 1.8 M H3PO4 initially. It will take 50 mL of 1.8 M KOH for the first eq pt so the concn of the salt will be 1.8 x (50/100) = 0.09
Post your work if you get stuck.