Calculate the ph of a 0.50 m hio. the ka of hypoiodic acid, hio, is 2.3x10–11.0.305.325.479.474.80

pH = 5.47

Explanation:

The equilibrium that takes place is:

HIO ↔ H⁺ + IO⁻

Ka = = 2.3 * 10⁻¹¹

At equilibrium:

Replacing those values in the equation for Ka and solving for x:

Then [H⁺]=3.39 * 10⁻⁶, thus pH = 5.47

you need to take a look at the periodic system in order to sort these elements into pairs that would most likely exhibit similar chemical properties. first, you need to find elements that are found in the same group (groups are vertical in the periodic system), and then to see whether they have the same valence number. if they meet these criteria, they will have similar properties.

the first pair: ne (neon) and kr (krypton), are noble gases found in group 8

the second pair: li (lithium) and k (potassium), found in group 1

the third pair: s (sulfur) and te (tellurium), found in group 6