The half-equivalence point of a titration occurs half way to the end point, where half of the analyte has reacted to form its conjugate, and the other half still remains unreacted. if 0.500 moles of a monoprotic weak acid (ka = 5.0 × 10-5) is titrated with naoh, what is the ph of the solution at the half-equivalence point?

Answer is: pH of the solution at the half-equivalence point is 4,30.
Chemical reaction: HA + NaOH → NaA + H₂O.
n₀(HA) = 0,500 mol.
Ka = 5,0·10⁻⁵.
Half of the analyte has reacted, so n(HA) = 0,5 mol ÷ 2 = 0,25 mol.
From chemical reaction: n(HA) : n(NaOH) = 1 : 1.
n(NaOH) = n(NaA) = 0,25 mol.
pH = pKa + log(n(NaA)/n(HA).
pH = -log(5,0·10⁻⁵) + log (0,25 mol/0,25 mol).
pH = 4,30.

The density of water is 1 g/cm cubed and in order for the pebble to sink its density has to be greater than the density of water and in order for the twig to float its density has to be less than the density of water