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Anuta_ua [19.1K]
3 years ago
9

How much 10.0 M HNO3 must be added to 1.00 L of a buffer that is 0.0100 M acetic acid and 0.100 M sodium acetate to reduce the p

H to 5.05 ?
Chemistry
1 answer:
sladkih [1.3K]3 years ago
8 0
Initial moles of CH₃COOH = 1.00 x 0.0100 = 0.0100 mol

Initial moles of CH₃COONa = 1.00 x 0.100 = 0.100 mol


Let a be the volume of HNO₃ that must be added

Moles of HNO₃ added = a/1000 x 10.0 = 0.01a mol

CH₃COONa + HNO₃ => CH₃COOH + NaNO₃


Moles of CH₃COOH = 0.0100 + 0.01a

Moles of CH₃COONa = 0.100 - 0.01a


pH = pKa + log([CH₃COONa]/[CH₃COOH])

= pKa + log(moles of CH₃COONa/moles of CH₃COOH)


5.05 = 4.75 + log((0.100 - 0.01a)/(0.0100 + 0.01a))

log((0.100 - 0.01a)/(0.0100 + 0.01a)) = 0.3

(0.100 - 0.01a)/(0.0100 + 0.01a) = 10^0.3 = 2.0

0.03 a = 0.08

a = 2.67


<span>Volume of HNO3 = a = </span>2.13 mL
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Both red and white phosphorus is made of phosphor element but they have a different structure which causes them to have a different color. This phenomenon called allotropes when a chemical element has two or more different form.They are not compounds as they are only made of phosphorus. A compound should be made by at least two different elements.
4 0
3 years ago
How do you know the ph level of 10^-8
sergiy2304 [10]

pH=6.98

Explanation:

This is a very interesting question because it tests your understanding of what it means to have a dynamic equilibrium going on in solution.

As you know, pure water undergoes self-ionization to form hydronium ions, H3O+, and hydroxide anions, OH−.

2H2O(l]⇌H3O+(aq]+OH−(aq]→ very important!

At room temperature, the value of water's ionization constant, KW, is equal to 10−14. This means that you have

KW=[H3O+]⋅[OH−]=10−14

Since the concentrations of hydronium and hydroxide ions are equal for pure water, you will have

[H3O+]=√10−14=10−7M

The pH of pure water will thus be

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5 0
3 years ago
The diameter of the He He atom is approximately 0.10 nm nm . Calculate the density of the He atom in g/cm 3 g/cm3 (assuming that
Sladkaya [172]

Answer:

Density of the He atom = 12.69 g/cm³

Explanation:

From the information given:

Since 1 mole of an atom = 6.022x 10²³ atoms)

1 atom of He = 1  \ atom \times  (\dfrac{1  \ mole}{  6.022 \times  10^{23}  \ atoms}) \times ( \dfrac{4.003 \ grams}{  1  \ mole})

=6.647 \times  10^{-24} \  grams

The volume can be determined as  folows:

since the diameter of the He atom is approximately 0.10 nm

the radius of the He = \dfrac{0.10}{2} = 0.05 nm

Converting it into cm, we have:

0.05 nm \times  \dfrac{10^{-9} \  meters}{ 1  nm} \times \dfrac{ 1 cm }{10^{-2} \ meters}

=5 \times  10^{-9}  \ cm

Assuming that it is a sphere, the volume of a sphere is

= \dfrac{4}{3}\pi r^3

= \dfrac{4}{3}\pi  \times (5\times 10^{-9})^3

= 5.236 \times 10^{-25} \ cm^3

Finally, the density can be calcuated by using the formula :

Density = \dfrac{mass}{volume}

D =  \dfrac{6.647 \times  10^{-24} \  grams }{ 5.236 \times 10^{-25} \  cm^3}

D = 12.69 g/cm³

Density of the He atom = 12.69 g/cm³

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