Determine the poh of each solution.
1. [h3o+] = 3.3×10−10 m
express your answer using two decimal places.
2. [h3o+] = 2.6×10−7 m
express your answer using two decimal places.
3. [h3o+] = 5.3×10−2 m
express your answer using two decimal places.
4. [oh−] = 1.32×10−4 m
express your answer using three decimal places.

1. pOH = 5.519; 2. pOH = 6.585; 3. pOH = 12.725; 4. pOH = 3.879

Explanation:

First of all, rembember this: water's equilibrium

2H₂O ⇄ H₃O⁺  +  OH⁻

The expression for Kw = [H₃O⁺] . [OH⁻] → 1x10⁻¹⁴

[H₃O⁺] = 1x10⁻¹⁴ / [OH⁻]

[OH⁻] = 1x10⁻¹⁴ / [H₃O⁺]

-log [H₃O⁺] = pH

-log [OH⁻] = pOH

pH + pOH = 14 because pKw = 14

1. -log 3.3×10⁻¹⁰ = 9.481

14 - 9.481 = 5.519

2. -log 2.6×10⁻⁷ = 6.585

14 - 6.585 = 7.415

3. -log 5.3×10⁻² = 1.275

14 - 1.275 = 12.725

4. - log  1.32×10⁻⁴ = 3.879

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

the answer is b. dilution