Answer:
0.0052. 1.3. 4.0. 6.7. The second EQ point
Explanation:
We can use the [tex]pK_{a1}[/tex] and [tex]pK_{a2}[/tex] to estimate the values of [tex]K_{a1}[/tex] and [tex]K_{a2}[/tex]. Therefore:
[tex]K_{a1}[/tex] = [tex]10^{-1.3}[/tex] = 0.0501
[tex]K_{a2}[/tex] = [tex]10^{-6.7}[/tex] = 1.995*10^(-7)
The value of [tex]K_{a1}[/tex] is largely more than [tex]pK_{a2}[/tex] thus, the effects of second ionization and [tex]K_{a2}[/tex] can be ignored. Therefore:
[tex]H_{3}PO_{3}[/tex] ⇒ [tex]H^{+} + H_{2}PO^{-} _{3}[/tex]
I 1.8 0 0
C -x x x
E 1.8-x x x
[tex]K_{a1}[/tex] = [tex]\frac{[H^{+}]*[H_{2}PO^{-} _{3}]}{[H_{3}PO_{3}]}[/tex]
1.3 = x*x/(1.8-x)
x^2 + 1.3x - 2.34 = 0
using x = [±b - sqrt(b^2 -4ac)]/2a
a = 1, b = 1.3, and c = -2.34
x = [-1.3 ± sqrt(1.69 + 9.36)]/2 = (-1.3 ± 3.324)/2
Since x can only be positive, we have:
x = 1.012 M
So pH = 0.0052
When 25 ml KOH is added, the moles added is half the moles of [tex]H_{3}PO_{3}[/tex], therefore, pH is equivalent to pKa1 which is equal to 1.3
when 50 ml KOH is added, pH = (1.3 + 6.7)/2 = 4.0
when 75 ml KOH is added, pH is equivalent to pKa2 which is equal to 6.7
when 100 ml KOH is added, the number moles of [tex]OH^{-}[/tex] is two times the moles of [tex]H_{3}PO_{3}[/tex] and this corresponds to the second EQ point.
The pH for each of the points in the titration of 50.0 mL of 1.8 M H 3 PO 3 ( aq ) with 1.8 M KOH ( aq ) is 0.0052. 1.3. 4.0. 6.7 and the second EQ point.
We would make use of pKa1 and pKa2 to make an estimate of the values of Ka1 and Ka2.
Ka1= 0.0501
Ka2= 1.995*10^(-7)
1.3 = x*x/(1.8-x)
x^2 + 1.3x - 2.34 = 0
using x = [±b - sqrt(b^2 -4ac)]/2a
a = 1, b = 1.3, and c = -2.34
x = [-1.3 ± sqrt(1.69 + 9.36)]/2 = (-1.3 ± 3.324)/2
Hence,
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