I'm not positive but I believe you are talking about where sperm comes from, correct?
Answer:
Ka3 for the triprotic acid is 7.69*10^-11
Explanation:
Step 1: Data given
Ka1 = 0.0053
Ka2 = 1.5 * 10^-7
pH at the second equivalence point = 8.469
Step 2: Calculate Ka3
pKa = -log (Ka2) = 6.824
The pH at the second equivalence point (8.469) will be the average of pKa2 and pKa3. So,
8.469 = (6.824 + pKa3) / 2
pKa3 = 10.114
Ka3 = 10^-10.114 = 7.69*10^-11
Ka3 for the triprotic acid is 7.69*10^-11
An abyssal plain......................................
The first dissociation for H2X:
H2X +H2O ↔ HX + H3O
initial 0.15 0 0
change -X +X +X
at equlibrium 0.15-X X X
because Ka1 is small we can assume neglect x in H2X concentration
Ka1 = [HX][H3O]/[H2X]
4.5x10^-6 =( X )(X) / (0.15)
X = √(4.5x10^-6*0.15)
∴X = 8.2 x 10-4 m
∴[HX] & [H3O] = 8.2x10^-4
the second dissociation of H2X
HX + H2O↔ X^2 + H3O
8.2x10^-4 Y 8.2x10^-4
Ka2 for Hx = 1.2x10^-11
Ka2 = [X2][H3O]/[HX]
1.2x10^-11= y (8.2x10^-4)*(8.2x10^-4)
∴y = 1.78x10^-5
∴[X^2] = 1.78x10^-5 m
Answer:
Keq =1.50108
Explanation:
The given reactionis
C₂H₂(g) +2H₂(g) -------------> C₂H₂(g)
ΔG0 f=ΔG0f n (products) - ΔG0f n (reactants )
= -32.89 kJ/mol - (209.2 kJ/mol+2*0.0 kJ/mol)
= - 242.09kJ/mol
ΔG= -RTlnKeq
ln Keq = -ΔG/RT
=-(- 242.09kJ/mol ) / 2 k cal /mol*298 K
=0.406
Keq =e0.406
Keq =1.50108
