Calculate the pH of a solution with [H+]= 1 x 10^-6M

A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67 mL of 0.175 M HCl to reach the equivalenc

Lorico [155]

Answer:

a. 0.180M of C₆H₅NH₂

b. 0.0887M C₆H₅NH₃⁺

c. pH = 2.83

Explanation:

a. Based in the chemical equation:

C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)

1 mole of aniline reacts per mole of HCl

Moles required to reach equivalence point are:

Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂

As the original solution had a volume of 25.0mL = 0.0250L:

4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂

b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles

But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:

[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺

c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:

C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺

Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =

2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]

When the system reaches equilibrium, molar concentrations are:

[C₆H₅NH₃⁺] = 0.0887M - X

[C₆H₅NH₂] = X

[H₃O⁺] = X

Replacing in Ka formula:

2.5x10⁻⁵ = [X] [X] / [0.0887M - X]

2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²

0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶

Solving for X:

X = -0.0015 → False solution. There is no negative concentrations.

X = 0.001477 → Right solution.

As [H₃O⁺] = X, [H₃O⁺] = 0.001477

Knowing pH = -log [H₃O⁺]

pH = -log 0.001477

7 0

3 years ago

Identify the correct mole ratio for each substance. Sodium chloride (NaCl) Na:Cl = 1: Ammonium nitrate (NHNO) H:O = 4:

Norma-Jean [14]

Answer: The mole ratio of H : O in ammonium nitrate is 4 : 3.

Explanation:

We are given a compound named ammonium nitrate having formula $NH_4NO_3$

There are 3 elements in this compound which are nitrogen, hydrogen and oxygen.

To calculate the mole ratio, we write the ratio of their subscripts. For this compound, it is:

$N:H:O::2:4:3$

The mole ratio of H and O for this compound is 4 : 3.

4 0

3 years ago

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Answer:

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Explanation:

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Answer:

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