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Ira Lisetskai [31]
3 years ago
9

A 150.0 mL sample of 0.20 M HF is titrated with 0.10 M LiOH. Determine the pH of the solution after the addition of 600.0 mL of

LiOH. The Ka of HF is 6.8 × 10-4.
Chemistry
1 answer:
s2008m [1.1K]3 years ago
7 0

Answer:

pH = 12.6

Explanation:

The HF reacts with LiOH as follows:

HF + LiOH → LiF + H₂O

To solve this question we need to find the moles of each reactant:

<em>Moles HF:</em>

0.1500L * (0.20mol / L) = 0.030 moles HF

<em>Moles LiOH:</em>

0.600L * (0.10mol / L) = 0.060 moles LiOH

That means there is an amount of LiOH in excess, that is:

0.060 mol - 0.030 mol = 0.030 moles LiOH

In 600.0mL + 150.0mL = 750.0mL = 0.750L

The molarity of LiOH is:

0.030 moles LiOH / 0.750L =

0.040M LiOH = [OH⁻]

As:

Kw = 1x10⁻¹⁴ = [H⁺] [OH⁻]

1x10⁻¹⁴ = [H⁺] [0.040M]

2.5x10⁻¹³M = [H⁺]

As pH = -log [H⁺]

<h3>pH = 12.6</h3>
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Please help me
Wittaler [7]

Answer:

pH = 6.999

The solution is acidic.

Explanation:

HBr is a strong acid, a very strong one.

In water, this acid is totally dissociated.

HBr + H₂O  →  H₃O⁺  +  Br⁻

We can think pH, as - log 7.75×10⁻¹² but this is 11.1

acid pH can't never be higher than 7.

We apply the charge balance:

[H⁺] = [Br⁻] + [OH⁻]

All the protons come from the bromide and the OH⁻ that come from water.

We can also think [OH⁻] = Kw / [H⁺] so:

[H⁺] = [Br⁻] + Kw / [H⁺]

Now, our unknown is [H⁺]

[H⁺] =  7.75×10⁻¹² + 1×10⁻¹⁴ / [H⁺]

[H⁺] = (7.75×10⁻¹² [H⁺] + 1×10⁻¹⁴) /  [H⁺]

This is quadratic equation:  [H⁺]² - 7.75×10⁻¹² [H⁺] - 1×10⁻¹⁴

a = 1 ; b = - 7.75×10⁻¹² ; c = -1×10⁻¹⁴

(-b +- √(b² - 4ac) / (2a)

[H⁺] = 1.000038751×10⁻⁷

- log [H⁺] = pH → 6.999

A very strong acid as HBr, in this case, it is so diluted that its pH is almost neutral.

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