If the ka of a monoprotic weak acid is 1.0 × 10-6, what is the ph of a 0.50 m solution of this acid?

Answer is: pH value of weak acid is 3.15.
Chemical reaction: HA(aq) ⇄ A⁻(aq) + H⁺(aq).
c(weak acid) = 0,5 M.
Ka(weak acid) = 1.0·10⁻⁶.
[A⁻] = [H⁺] = x.
[HA] = 0.5 M - x.
Ka = [A⁻]·[H⁺] / [HA]. 
1.0·10⁻⁶ = x² / (0,5 M - x).
Solve quadratic equation: x = [H⁺] = 0.0007 M.
pH = -log[H⁺].
pH = -log(0.0007 M).
pH = 3.15.

Manganese is reduced from MnIV+ to MnII+, so option A.

Explanation:

And of course, for every reduction there is a corresponding oxidation: here, chloride ion is oxidized to zerovalent chlorine gas.

The overall (balanced) redox equation is:

MnO2(s)+4HCl(aq)→MnCl2(s)+Cl2(g)+2H2O(l)