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vagabundo [1.1K]
3 years ago
6

How many grams of NaF should be added to 500 mL of a 0.100 M solution of HF to make a buffer with a pH of 3.2

Chemistry
1 answer:
Korolek [52]3 years ago
4 0

Answer:

2.25g of NaF are needed to prepare the buffer of pH = 3.2

Explanation:

The mixture of a weak acid (HF) with its conjugate base (NaF), produce a buffer. To find the pH of a buffer we must use H-H equation:

pH = pKa + log [A-] / [HA]

<em>Where pH is the pH of the buffer that you want = 3.2, pKa is the pKa of HF = 3.17, and [] could be taken as the moles of A-, the conjugate base (NaF) and the weak acid, HA, (HF). </em>

The moles of HF are:

500mL = 0.500L * (0.100mol/L) = 0.0500 moles HF

Replacing:

3.2 = 3.17 + log [A-] / [0.0500moles]

0.03 = log [A-] / [0.0500moles]

1.017152 = [A-] / [0.0500moles]

[A-] = 0.0500mol * 1.017152

[A-] = 0.0536 moles NaF

The mass could be obtained using the molar mass of NaF (41.99g/mol):

0.0536 moles NaF * (41.99g/mol) =

<h3>2.25g of NaF are needed to prepare the buffer of pH = 3.2</h3>
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Explanation:

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Using ideal gas equation, we will calculate the number of moles as follows.

                                PV = nRT

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                                          = 1 \times \frac{0.720 L}{0.0821 \times 298}

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Let us assume that there are x mol of Ar and y mol of Xe.

Hence, total number of moles will be as follows.

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Also,      40x + 131y = 2.966

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Therefore, mole fraction will be calculated as follows.

Mol fraction of Xe = \frac{y}{(x+y)}

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