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What is the approximate value of the equilibrium constant, k n, for the neutralization of acetic acid withsodium hydroxide, show

NikAS [45]

You are given the neutralization of acetic acid with sodium hydroxide. Also, you are given the k for acetic acid, which is 1.8 x 10⁻⁵. You are asked to find the approximate value of the equilibrium constant, kn, for the neutralization. We will have a reaction of both acetic acid and sodium hydroxide.

CH₃COOH + NaOH → CH₃COONa + H₂O
which comes from
CH₃COOH → CH₃COO⁻ + H⁺
H⁺ + OH⁻ → H₂O

The k for water is always 1.0 x 10¹⁴. The Ksp for the reaction will be

Ksp = [CH₃COOH][H₂O]
Ksp = (1.8 x 10⁻⁵)(1.0 x 10¹⁴)
Ksp = 1.8 x 10⁹

5 0

3 years ago

If 12.7 grams of hydrogen reacted with an unknown amount of oxygen, how many grams of water would be produced? (Write a balanced

nirvana33 [79]

Answer:

48 g.

Explanation:

7 0

2 years ago

Read 2 more answers

Which of the following materials will conduct electricity well?

Alborosie

Answer:

Metal paper clips hope it helps

7 0

3 years ago

How much energy is required to convert 15.0 g of ice at −106 °C to water vapor at 125 °C? Specific heats are 2.09 J/g K for both

myrzilka [38]

Answer:

49.3 kJ of energy is required

Explanation:

An exercise of calorimetry at its best

First of all, convert the ice to water before melting.

Q = ice mass . C . ΔT

Q = 15 g . 2.09 J/g°C (0° - (-106°C)

15 g . 2.09 J/g°C . 106°C = 3323.1 J

Now we have to melt the ice, to change its state

Q = mass . latent heat of fusion

Q = 15 g . 0.335 kJ/g = 5.025 kJ .1000 = 5025 J

After that, we have liquid water at 0° and the ice has melted completely. We have to release energy to make a temperature change, to 100° (vaporization)

Q = 15g . 4.18 J/g°C (100°C - 0°C)

Q = 6270 J

Water has been vaporizated so we have to calculate, the state change.

Q = mass . latent heat of vap

Q = 15 g. 2.260 kJ/g

Q = 33.9 kJ (.1000) = 33900 J

Finally we have to increase temperature from 100°C to 125°C

Q = 15 g . 2.09 J/g°C . (125°C - 100°C)

Q = 783.75 J

To know how much energy is required to conver 15 g of ice, to water vapor at 125°C, just sum all the heat released.

3323.1 J + 5025 J + 6270 J + 33900 J + 783.75 J = 49301.85 joules.

Notice I have to convert kJ to J in two calcules to make the sum.

49301.85 joules / 1000 = 49.3 kJ

4 0

3 years ago

Buffer preparation. You wish to prepare a buffer consisting of acetic acid and sodium acetate with a total acetic acid plus acet

Hoochie [10]

Answer:

0.182 moles of acetic acid are needed, this means 10.93 g.

0.318 moles of sodium acetate are needed, this means 26.08 g.

Explanation:

The Henderson–Hasselbalch (H-H) equation tells us the relationship between the concentration of an acid, its conjugate base, and the pH of a buffer:

pH = pka + $log\frac{[A^{-} ]}{[HA]}$

In this case, [A⁻] is the concentration of sodium acetate, and [HA] is the concentration of acetic acid. The pka is a value that can be looked up in literature: 4.76.

From the problem we know that

[A⁻] + [HA] = 250 mM = 0.250 M eq. 1

We use the H-H equation, using the data we know, to describe [A⁻] in terms of [HA]:

5.0 = 4.76 + $log\frac{[A^{-} ]}{[HA]}$

$0.24=log\frac{[A^{-} ]}{[HA]}\\\\10^{0.24}=\frac{[A^{-} ]}{[HA]}\\ 1.74 [HA] = [A^{-}]$ eq.2

Now we replace the value of [A⁻] in eq. 1, to calculate [HA]:

1.74 [HA] + [HA] = 0.250 M

[HA] = 0.091 M

Then we calculate [A⁻]:

[A⁻] + 0.091 M = 0.250 M

[A⁻] = 0.159 M

Using the volume, we can calculate the moles of each substance:

moles of acetic acid = 0.091 M * 2 L = 0.182 moles
moles of sodium acetate = 0.159 M * 2 L = 0.318 moles

Using the molecular weight, we can calculate the grams of each substance:

grams of acetic acid = 0.182 mol * 60.05 g/mol = 10.93 g
grams of sodium acetate = 0.318 mol * 82.03 g/mol = 26.08 g

8 0

3 years ago