7.86 is the pOH of water at this temperature of 100 degrees celsius.
Option E is the right answer.
Explanation:
Data given:
Kw = 51.3 x 
pOH = ?
we know that pure water is neutral and will have pH pf 7.
The equation for relation between Kw and H+ and OH- ion is given by:
Kw = [H+] [OH-}
here the concentration of H+ ion and OH- ion is equal
so, [H+]= [OH-]
Putting the values in the equation of Kw
pKw = -log[Kw]
pKw = -log [51.3 x
]
pKw = 12.28
since H+ ion OH ion concentration is equal the pH of water is half i.e. 6.14
Now, pOH is calculated by using the equation:
14 = pOH + pH
14- 6.14 = pOH
pOH = 7.86
Reaction of dissociation: Ag₂SO₄ → 2Ag⁺ + SO₄²⁻.
m(Ag₂SO₄) = 4 g.
V(Ag₂SO₄) = 1 l.
n(Ag₂SO₄) = m(Ag₂SO₄) ÷ M(Ag₂SO₄).
n(Ag₂SO₄) = 4 g ÷ 311,8 g/mol.
n(Ag₂SO₄) = 0,0128 mol.
n(Ag⁺) = 2 · 0,0128 mol = 0,0256 mol.
n(Ag₂SO₄) = n(SO₄²⁻) = 0,0128 mol.
c(Ag⁺) = n ÷ V = 0,0256 mol ÷ 1 l = 0,0256 mol/l.
Ksp = c(Ag⁺)² · c(SO₄²⁻).
Ksp = (0,0256 mol/l)² · 0,0128 mol/l.
Ksp = 8,3·10⁻⁶.
Mass of the water : 2.23 g
<h3>Furter explanation</h3>
Heat
Q = m.c.Δt
m= mass, g
c = heat capacity, for water : 4.18 J/g° C.
ΔT = temperature
Q= 140 J
Δt = 75 - 60 = 15
mass of the water :

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