Calculate the hydrogen-ion concentration [H+] for the aqueous solution in which [OH–] is 1 x 10–4 mol/L. Is this solution acidic, basic, or neutral? Show your work.

[H⁺] = 1×10⁻¹⁰     →  Basic solution

Explanation:

For this question we need to know, the autoionization of water:

2H₂O (l)  ⇄  H₃O⁺ (aq) + OH⁻(aq)     Kw

Kw = 1ₓ10⁻¹⁴ → [H₃O⁺] . [OH⁻]

1ₓ10⁻¹⁴ /[OH⁻] = [H₃O⁺] → 1ₓ10⁻¹⁴ / 1×10⁻⁴  =  1×10⁻¹⁰

In order to determine if the solution is acidic, basic or neutral we use the pH value.

pH = - log  [H₃O⁺]  → -log 1ₓ10⁻¹⁰  = pH = 10          Basic solution

pH < 7 → Acidic solution

pH = 7 → Neutral solution

pH > 7 → Basic solution

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g