What is a 1 unit in ph equal to

Chemistry

1 answer:

notka56 [123]3 years ago

8 0

Answer:

Since pH is a logarithmic scale, a difference of one pH unit is equivalent to a tenfold difference in hydrogen ion concentration. ... Pure water and a solution of NaCl in pure water are both neutral, since dissociation of water produces equal numbers of both ions.

Explanation:

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In sediments and waterlogged soil, dissolved O2 concentrations are so low that the microorganisms living there must rely on othe

kakasveta [241]

Answer:

1) SO₄ ²⁻ : (+6)

H₂S : (-2)

Explanation:

a) <u>Sulfate reducers</u> are widespread in muds and other sediments, water-logged soils, etc., environments that contain SO₄ ²⁻ and become anoxic as a result of microbial decomposition.

Sulfate (SO₄ ²⁻), the most oxidized form of sulfur (+6), <u>is reduced</u> by these

sulfate-reducing bacteria. The end product of sulfate reduction is hydrogen sulfide, H₂S, (oxidation number -2) an important natural product that participates in many biogeochemical processes. The H₂S they generate is responsible for the pungent smell (like that of rotten eggs) often encountered near coastal ecosystems. When sulfate-reducing bacteria grow, the H₂S formed from SO₄ ²⁻ reduction combines with the ferrous iron to form black, insoluble ferrous sulfide, which is not toxic. This is important for the conservation of the environment.

b) The net ionic equation under acidic conditions is:

4 H₂ + SO₄²⁻ + H⁺ → HS⁻ + 4 H₂O

Global reaction: SO₄²⁻ + 2H⁺ → H₂S + O₂

3 0

2 years ago

G determine the concentration of an hbr solution if a 45.00 ml aliquot of the solution yields 0.6485 g agbr when added to a solu

Sunny_sXe [5.5K]

The molecular weight of silver bromide (AgBr) is 187.77 g/mole. The presence of the ions in solution can be shown as- AgBr (insoluble) ⇄ $Ag^{+}$ + $Br^{-1}$ .

45.00 mL of the aliquot contains 0.6485 g of AgBr. Thus 1000 mL of the aliquot contains $\frac{0.6485}{45}$ ×1000 = 14.411 gm-mole. Thus the solubility product $K_{sp}$ of AgBr = [ $Ag^{+}$ ]× $Br^{-}$ .