Not sure how to do this problem:
Determine the pH of a 1L solution of 0.2M phosphoric acid to which 0.5 mol of KOH is added.
The conjugate base I know is H2PO4-.
0.5 mol KOH is in a 1L solution so the concentration is 0.5 mol/L.
So since KOH is added to the 1L solution of 0.2M phosphoric acid, I added 0.5M KOH to 0.2M phosphoric acid, so that the new concentration of phosphoric acid is 0.7M.
"x" is H2PO4- because I was trying to find out what the concentration was.
I used the Ka of H3PO4 which was 7.08x10^-3 (was I supposed to use the Ka of H2PjO4?)
I did Ka = [A-]/[HA]
7.03x10^-3 = x/[0.7M]
(7.03x10^-3)(0.7M) = x
x = 4.956x10^-3M = [H2PO4-]
Then I used the Henderson Hasselbalch equation: pH = pKa+log[A-]/[HA]
I used the pKa of H2PO4 which was 6.82.
pH = 6.82 + log [4.956x10^-3/0.7] = 4.67
Is this right? I don't know when to use a certain pKa or Ka value. For example, phosphoric acid has 3 pKas. When do I use each? I just decided to use the pKa for H2PO4 because it had a better answer than when I used the pK for H3PO4 which was 2.15. Same thing goes for Ka - not sure which one to use and when.
Thanks in advance.
Determining pH of a solution - addition of KOH to phosphoric acid. Did I solve this problem correctly?
2016-10-16 5:49 am
回答 (1)
2016-10-16 12:27 pm
Your answers are incorrect. This is because H₃PO₄ completely reacts.
Initial number of moles of H₃PO₄ = (0.2 mol/L) × (1 L) = 0.2 mol
Initial number of moles of KOH = 0.5 mol > 0.2 mol
H₃PO₄(aq) + KOH(aq) ⇌ KH₂PO₄(aq) + H₂O(l)
0.2 mol of KOH reacts with 0.2 mol H₃PO₄ to give 0.2 mol of KH₂PO₄(aq).
No. of moles of KOH after above reaction = (0.5 - 0.2) mol = 0.1 mol
KH₂PO₄(aq) + KOH(aq) ⇌ K₂HPO₄(aq) + H₂O(l)
0.2 mol of KOH reacts with 0.2 mol KH₂PO₄ to give 0.2 mol of K₂HPO₄(aq).
No. of moles of KOH after above reaction = (0.3 - 0.1) mol = 0.1 mol
K₂HPO₄(aq) + KOH(aq) ⇌ K₃PO₄(aq) + H₂O(l)
0.1 mol of KOH reacts with 0.1 mol K₂HPO₄ to give 0.1 mol of K₃PO₄(aq).
No. of moles of KOH after above reaction = (0.1 - 0.1) mol = 0 mol
In the final solution :
[HPO₄²⁻] = (0.2 - 0.1) M = 0.1 M
[PO₄³⁻] = 0.1 M
Consider the dissociation of HPO₄²⁻ ions :
HPO₄²⁻(aq) + H₂O(l) ⇌ PO₄³⁻(aq) + H₃O⁺(aq) …… K₃ = 4.2 × 10⁻¹³
pH = pK₃ - log([HPO₄²⁻]/[PO₄³⁻]) = -log(4.2 × 10⁻¹³) - log(0.1/0.1) = 12.4
Initial number of moles of H₃PO₄ = (0.2 mol/L) × (1 L) = 0.2 mol
Initial number of moles of KOH = 0.5 mol > 0.2 mol
H₃PO₄(aq) + KOH(aq) ⇌ KH₂PO₄(aq) + H₂O(l)
0.2 mol of KOH reacts with 0.2 mol H₃PO₄ to give 0.2 mol of KH₂PO₄(aq).
No. of moles of KOH after above reaction = (0.5 - 0.2) mol = 0.1 mol
KH₂PO₄(aq) + KOH(aq) ⇌ K₂HPO₄(aq) + H₂O(l)
0.2 mol of KOH reacts with 0.2 mol KH₂PO₄ to give 0.2 mol of K₂HPO₄(aq).
No. of moles of KOH after above reaction = (0.3 - 0.1) mol = 0.1 mol
K₂HPO₄(aq) + KOH(aq) ⇌ K₃PO₄(aq) + H₂O(l)
0.1 mol of KOH reacts with 0.1 mol K₂HPO₄ to give 0.1 mol of K₃PO₄(aq).
No. of moles of KOH after above reaction = (0.1 - 0.1) mol = 0 mol
In the final solution :
[HPO₄²⁻] = (0.2 - 0.1) M = 0.1 M
[PO₄³⁻] = 0.1 M
Consider the dissociation of HPO₄²⁻ ions :
HPO₄²⁻(aq) + H₂O(l) ⇌ PO₄³⁻(aq) + H₃O⁺(aq) …… K₃ = 4.2 × 10⁻¹³
pH = pK₃ - log([HPO₄²⁻]/[PO₄³⁻]) = -log(4.2 × 10⁻¹³) - log(0.1/0.1) = 12.4
收錄日期: 2021-04-18 15:41:29
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