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adelina 88 [10]
3 years ago
8

A value of pH 7.0 indicates A. an zero hydrogen-ion concentration. B. the lowest hydrogen-ion concentration. C. the highest hydr

ogen-ion concentration. D. equal hydrogen and hydroxide-ion concentration.
Chemistry
1 answer:
rodikova [14]3 years ago
3 0

Answer:

The correct answer is D. equal hydrogen and hydroxide-ion concentration.

Explanation:

The pH indicates the alkalinity or acidity of a solution. The pH less than 7 are acidic (more H + ions) and greater than 7 are basic (less H + ions). A pH of 7 indicates a neutral solution, being the case of water.

The water dissociation constant (Kw) = 1e10-14 = (H +) (OH-), and with pH = 7.00, the concentration of H + = antilog -pH = antilog -7 = 1e10-7 is calculated. Thus using the formula of kw, we calculate the concentration of OH- that at pH = 7, 00 is equal to the concentration of H +:

Kw = (H +) (OH-)

1e10-14 = 1e10-7 (OH-)

(OH -) = 1e10-14 / 1e10-7 = 1e10-7

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You are given a solid that is a mixture of Na2SO4 and K2SO4. A 0.205-g sample of the mixture is dissolved in water. An excess of
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Mass of SO₄⁻² = 0.123 g.

Mass percentage of SO₄⁻² = 41.2%

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Mass of K₂SO₄ = 0.1277 g

Explanation:

Here we have

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K₂SO₄ = Y

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Therefore since the BaSO₄ is formed from BaCl₂, Na₂SO₄ and K₂SO₄ we have

Amount of BaSO₄ from Na₂SO₄ is therefore;

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4}

Amount of BaSO₄ from K₂SO₄ is;

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4}

Molar mass of

BaSO₄ = 233.38 g/mol

Na₂SO₄ = 142.04 g/mol

K₂SO₄ = 174.259 g/mol

X\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, Na_2SO_4} = X\times\frac{233.38 }{142.04} = 1.643·X

Y\times\frac{Molar \, Mass \, of \, BaSO_4}{Molar \, Mass \, of \, K_2SO_4} = Y\times\frac{233.38 }{174.259 } = 1.339·Y

Therefore, we have

1.643·X + 1.339·Y = 0.298 g.....(2)

Solving equations (1) and (2) gives

The mass of SO₄⁻² in the sample is given by

Mass of sample = 0.298

Molar mass of BaSO₄ = 233.38 g/mol

Mass of Ba = 137.327 g/mol

∴ Mass of SO₄ = 233.38 g - 137.327 g = 96.05 g

Mass fraction of SO₄⁻² in BaSO₄ = 96.05 g/233.38 g = 0.412

Mass of SO₄⁻² in the sample is 0.412×0.298 = 0.123 g.

Percentage mass of SO₄⁻² = 41.2%

Solving equations (1) and (2) gives

X = 0.0773 g and Y = 0.1277 g.

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