Hclo is a weak acid (ka = 4.0 × 10–8) and so the salt naclo acts as a weak base. what is the ph of a solution that is 0.030 m in naclo at 25 °c?

Chemical reaction 1: NaClO(aq) → Na⁺(aq) + ClO⁻(aq).
Chemical reaction 2: ClO⁻(aq) + H₂O(l) ⇄ HClO(aq)+ OH⁻(aq).
c(NaClO) = 0,030 M.
[ClO⁻] = 0,03 M - x.
Ka(HClO) = 4·10⁻⁸.
Ka · Kb = 10⁻¹⁴.
Kb(ClO⁻) = 2,5·10⁻⁷.
Kb(ClO⁻) = [OH⁻] · [HClO] / [ClO⁻].
[OH⁻] = [HClO] = x.
2,5·10⁻⁷ = x² / (0,03 M -x).
Solve quadratic equation: x = [OH⁻] = 0.0000893 M.
pOH = -log(0.0000893 M) = 4.05.
pH = 14 - 4.05.
pH = 9.95.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.