We assume that we have Ka= 4.2x10^-13 (missing in the question)
and when we have this equation:
H2PO4 (-) → H+ + HPO4-
and form the Ka equation we can get [H+]:
Ka= [H+] [HPO4-] / [H2PO4] and we have Ka= 4.2x10^-13 & [H2PO4-] = 0.55m
by substitution:
4.2x10^-13 = (z)(z)/ 0.55
z^2 = 2.31x 10^-13
z= 4.81x10^-7
∴[H+] = 4.81x10^-7
when PH equation is:
PH= -㏒[H+]
= -㏒(4.81x10^-7) = 6.32
Answer:
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Answer: -
100 mm Hg
Explanation: -
P 1 =400 mm Hg
T 1 = 63.5 C + 273 = 336.5 K
T 2 = 34.9 C + 273 = 307.9 K
ΔHvap = 39.3 KJ/mol = 39.3 x 10³ J mol⁻¹
R = 8.314 J ⁻¹K mol⁻¹
Now using the Clausius Clapeyron equation
ln (P1 / P2) = ΔHvap / R x (1 / T2 - 1 / T1)
Plugging in the values
ln (400 mm/ P₂) = (39.3 x 10³ J mol⁻¹ / 8.314 J ⁻¹K mol⁻¹) x (
- 
= 1.38
P₂ = 100 mm Hg