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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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3 years ago
How many grams of \text{NaCl}NaClstart text, N, a, C, l, end text will be produced from 18.0 \text{ g}18.0 g18, point, 0, start
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Answer:

18.7887 g of NaCl

Explanation:

<em>The question reads - How many grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2?</em>

Let us start by writing out the balanced equation of the reaction:

Na + Cl2 ---> NaCl2

1 mole, each of Na and Cl2 is required to produce 1 mole of NaCl.

mole = mass/molar mass

Therefore

18 g of Na = 18/23 = 0.7826 mole

23 g of Cl2 = 23/71 = 0.3239 mole

In this case, the Na is in excess and the Cl2 becomes the limiting reagent. Hence

0.3239 mole of Cl2 will react with 0.3239 mole of Na to yield 0.3239 mole of NaCl.

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<u>Hence, 18.7887  grams of NaCl will be produced from 18.0 g of Na and 23.0 g of Cl2.</u>

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3 years ago
How many moles are in 5.67x10^24 atoms of RbCl?
Mademuasel [1]

Answer:

<h2>9.42 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.67 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 9.418604...

We have the final answer as

<h3>9.42 moles</h3>

Hope this helps you

4 0
3 years ago
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