When we double the mass the amount of KE is

Calculate the pH of 1.00 L of the buffer 0.95 M CH3COONa/0.92 M CH3COOH before and after the addition of the following species.

natima [27]

Answer: (a) pH = 4.774, (b) pH = 4.811 and (c) pH = 4.681

Explanation: (a) pH of the buffer solution is calculated using Handerson equation:

$pH=pKa+log(\frac{base}{acid})$

pKa for acetic acid is 4.76. concentration of base and acid are given as 0.95M and 0.92M. Let's plug in the values in the equation and calculate the pH of starting buffer.

$pH=4.76+log(\frac{0.95}{0.92})$

pH = 4.76 + 0.014

pH = 4.774

(b) When 0.040 mol of NaOH (strong base) are added to the buffer then it reacts with 0.040 mol of acetic acid and form 0.040 mol of sodium acetate.

Original buffer volume is 1.00 L. So, the original moles of sodium acetate will be 0.95 and acetic acid will be 0.92.

moles of acetic acid after addition of NaOH = 0.92 - 0.040 = 0.88

moles of sodium acetate after addition of NaOH = 0.95 + 0.040 = 0.99

Let's again plug in the values in the Handerson equation:

$pH=4.76+log(\frac{0.99}{0.88})$

pH = 4.76 + 0.051

pH = 4.811

(c) When 0.100 mol of HCl are added then it reacts with exactly 0.100 moles of sodium acetate(base) and form 0.100 moles of acetic acid(acid).

so, new moles of acetic acid = 0.92 + 0.100 = 1.02

new moles of sodium acetate = 0.95 - 0.100 = 0.85

Let's plug in the values in the equation:

$pH=4.76+log(\frac{0.85}{1.02})$

pH = 4.76 - 0.079

pH = 4.681

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4 years ago

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valentina_108 [34]

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3 years ago

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Which molecule will undergo only london dispersion forces when interacting with other molecules of the same kind?.

Neko [114]

Molecules undergo London dispersion forces:

$C_{4}H_{10}$ is the molecule will undergo only London dispersion forces when interacting with other molecules of the same kind.

What are London dispersion forces?

A sort of force that interacts between atoms and molecules that is often electrically symmetric is referred to as a London dispersion force.
When viewed from the nucleus, their electron distribution is frequently symmetrical. This dispersion force, which is also known as a transient attractive force, is frequently observed when the locations of the electrons in two nearby atoms cause the atoms to temporarily form dipoles.
The bond is polar when there are significant variations between the elements' electronegativities; it is nonpolar when there are similarities. When the molecule's dipole moment is equal to O, it is nonpolar; when it differs from O, it is polar.
The force at these molecules is known as the London dispersion force. In nonpolar molecules, the forces are weak, and partial charges must be induced so that they can bond. In polar molecules, partial charges caused by polarity result in a stronger link known as a dipole-dipole. The dipole-dipole is significantly stronger and known as a hydrogen bond if it is connected to a large electronegative atom (F, O, or N). Ionic force is the name for the attraction force at ionic substances.
The intermolecular force in the letter an is the London dispersion force because the compound is nonpolar;

Reason for incorrect options:

b: the compound is ionic because Na is a metal and the other part is covalent,

c: two compounds are possible: one is nonpolar and exhibits London dispersion force; the other is polar and exhibits dipole-dipole force; and

d: both compounds exhibit hydrogen bonds (H bonded to O, and H bonded to F).

NOTE: Your question is incomplete, but most probably your full question was, which molecule will undergo only London dispersion forces when interacting with other molecules of the same kind? Which molecule will undergo only London dispersion forces when interacting with other molecules of the same kind?

A. $C_{4}H_{10}$

B. $NaC_{2}H_{3}O_{2}$