(a) Since Ka1 >> Ka2, you can ignore the second ionization. However, you\\\'ll n
ID: 889299 • Letter: #
Question
(a) Since Ka1 >> Ka2, you can ignore the second ionization. However, you'll need to use either the quadratic equation or successive approximations to solve for [H ].
(b) 25.0 mL is half way to the first equivalence point, so pH = pKa1.
(c) 50.0 mL is the first equivalence point. H2PO3– is amphoteric, so use pH = 1/2(pKa1 pKa2).
(d) 75.0 mL is half way to the second equivalence point, so pH = pKa2.
(e) 100.0 mL is the second equivalence point. The final product, HPO32–, is a weak base with Kb1 = Kw/Ka2.
Explanation / Answer
Solution :-
a) Calculating pH before adding any KOH
Pka 1 = 1.30
Therefore Ka1 = Anitlog[-1.30] = 5.01*10^-2
Using the ka1 lets calculate the pH
H3PO3 + H2O ----- > H3O+ + H2PO3^-
3.0 M 0 0
-x +x +x
3.0-x x x
Ka1 = [H3O+] [H2PO3-]/[H3PO3]
5.01*10^-2 = [x][x]/[3.0-x]
5.01*10^-2 * 3.0-x = x^2
By solving this using the quadratic equation we get
X= 0.3634 M
Therefore pH is
pH= -log [H3O+]
pH = - log [0.3634]
pH= 0.440
b) calculating pH after adding 25 ml of 3.0 M KOH
molarity of the Acid is 3.0 M
therefore it reaches the half way of the first equivalence point
where moles of the acid and the moles of the conjugate base are same
therefore
pH= pka1
pH= 1.30
c) Calculating pH after adding 50 ml of 3.0 M KOH
At this addition titration reaches to the first eqeuivalence point means all the acid H3PO3 is converted to the H2PO3-
Therefore
pH = ½ (pka 1 + pka2)
pH = ½ (1.30+6.70)
pH= 4
d) Calculating pH after adding 75 ml of 3.0 M KOH
Here titration reaches to the half way of the second equivalence point means
Half of HPO3^- changes to PO3^2-
Therefore
pH = pka2
pH= 6.70
e) Calculating pH after adding 100 ml of 3.0 M KOH
At this point all the HPO3- changes to the PO3^2-
Therefore
Concentration of the PO3^2- is halved because the volume is doubled (100 ml + 100 ml =200 ml )
Therefore new concentration of the PO3^2- = 3.0 M / 2 = 1.5 M
Kb1= Kw / ka2
Kb1 = 1*10^-14 / 5.01*10^-2
Kb1 = 2.0*10^-13
Therefore using the Kb1 lets calculate the OH^-
PO3^2- + H2O -------- > HPO3^- + OH^-
1.5 M 0 0
-x +x +x
1.5-x x x
Kb1 = [HPO3^-][OH-]/[PO3^2-]
2.0*10^-13 = [x][x]/[1.5-x]
Sinc kb1 is very small therefore neglect the x from the denominator
Therefore
2.0*10^-13 = x^2 / 1.5
2.0*10^-13 * 1.5 = x^2
3*10^-13 =x^2
Taking square root of both sides we get
5.477*10^-7 = x = [OH-]
Now lets find the pOH
pOH= -log [OH-]
pOH= -log[5.477*10^-7]
pOH = 6.26
now lets find the pH
pH= 14 – pOH
pH= 14 – 6.26
pH= 7.74
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