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balu736 [363]
3 years ago
6

An unknown weak acid with a concentration of 0.086 M has a pH of 1.80. What is the Ka of the weak acid?

Chemistry
1 answer:
AlekseyPX3 years ago
5 0

Answer:

2.90 × 10⁻³

Explanation:

Step 1: Given data

pH: 1.80

Concentration of the weak acid (Ca): 0.086 M

Step 2: Calculate the concentration of H⁺

We will use the following expression.

pH = -log [H⁺]

[H⁺] = antilog -pH = antilog -1.80

[H⁺] = 0.0158 M

Step 3: Calculate the acid dissociation constant (Ka)

We will use the following expression.

Ka = [H⁺]² / Ca

Ka = 0.0158² / 0.086

Ka = 2.90 × 10⁻³

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Explanation:

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2 years ago
Write the balanced reaction and solubility product expression (KSP) for dissolving silver chromate: Ag2CrO4(s). Include all char
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Answer:

2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓

Ksp = [2s]²  . [s] → 4s³

Explanation:

Ag₂CrO₄ → 2Ag⁺  + CrO₄⁻²

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2Ag⁺ (aq)  + CrO₄⁻² (aq) ⇄  Ag₂CrO₄ (s) ↓ Ksp

That's the expression for the precipitation equilibrium.

To determine the solubility product expression, we work with the Ksp

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                          2 s                 s

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3 0
3 years ago
Calculate the mass of aluminum in 500 g of Al(C2H3O2)3
Luden [163]
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Explanation:

1) The ratio of Al in the molecule is  1 mol to 1 mol .

2) The mass of 1 mol of molecules of Al (CH2H3O2)3 is the molar mass of the compound.

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H: 3 * 3 * 1 g/mol = 9 g/mol
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4) Set a proportion:

    27 g/mol                x
-------------------- =  ----------
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5) Solve for x:

x = 500 g * 27 g/mol / 204 g/mol = 66.2 g
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Explanation:

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