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Mazyrski [523]
3 years ago
10

A student mixes 1.0 mL of aqueous silver nitrate, AgNO3 (aq), with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test

tube. What will the student observe
Chemistry
1 answer:
Lubov Fominskaja [6]3 years ago
4 0

Answer:

AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.

Explanation:

If a student mixes 1.0 mL of aqueous silver nitrate AgNO3 (aq)  with 1.0 mL of aqueous sodium chloride, NaCl (aq), in a clean test tube.

The sodium chloride is being acidified with dilute trioxonitrate (V) acid. Then a few drops of  silver trioxonitrate(V) is added afterwards. A  white precipitate of silver chloride, which dissolves readily in aqueous ammonia indicates the presence of sodium chloride.

The reaction proceeds as follows:

\mathtt{AgNO_{3(aq)} + NaCl _{(aq)} \to AgCl _{(s)} + NaNO_3_{(aq)}}

From the reaction between AgNO3 (aq) and NaCl (aq), AgCl (silver Chloride) is being precipitated out as white and cloudy crystals.

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How the calculation of the [OH-], pH and % ionization for 0.619 M ammonia (NH3) NH3 + H2O (liq) rightwards harpoon over leftward
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Answer:

[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52

Explanation:

Kb of the reaction:

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Is:

Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]

<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>

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<em>And as </em>[NH₃] = 0.619M

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1.11x10⁻⁵ = X²

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<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />

% ionization:

[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%

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As pOH = -log [OH-]

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<h3>pH = 11.52</h3>
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