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vovangra [49]
3 years ago
10

Prepare a solution that is 0.1 M acetic acid and 0.1 M sodium acetate by measuring out 5.0 mL of the 1.0 M acetic acid solution

and 5.0 mL of the 1.0 M sodium acetate solution in a 100 mL graduated cylinder, diluting the 10.0 mL to a final volume of 50.0 mL with deionized water, and then stirring. Pour this solution into a clean, dry 100 mL breaker. By knowing that the Ka for acetic acid is 1.8 x -5 10 , calculate the theoretical pH of the solution.
Chemistry
1 answer:
Juli2301 [7.4K]3 years ago
8 0

Answer:

4.74

Explanation:

It is possible to find pH of a buffer (The mixture of a weak acid: Acetic acid, with its conjugate base: Sodium acetate) using Henderson-Hasselbalch equation:

pH = pKa + log₁₀ [A⁻] / [HA]

<em>Where pKa is -log Ka of the weak acid,  [A⁻] concentration of the conjugate base and [HA] concentration of the weak acid</em>

pKa of acetic acid is -log 1.8x10⁻⁵ = 4.74

The concentration of both, acetic acid and sodium acetate is 0.1M. Replacing in H-H equation:

pH = pKa + log₁₀ [A⁻] / [HA]

pH = 4.74 + log₁₀ [0.1] / [0.1]

pH = 4.74 + log₁₀ 1

<em>pH = 4.74</em>

Theoretical pH is 4.74

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A student is driving her car when an insect strikes her windshield. Which statement correctly describes the forces in this situa
Scorpion4ik [409]

Answer:

The insect exerts no force on the windshield, and the windshield strikes the insect with a large force.

Explanation:

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6 0
3 years ago
Write the equilibrium constant expression for this reaction: CH3OH (aq)+Cl-(aq)-&gt; CH3Cl(aq)+OH-(aq)
bulgar [2K]

Answer: K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}

Explanation:

Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients. Pure solids are assumed to have a concentration of 1.

The given balanced equilibrium reaction is:

 CH_3OH(aq)+Cl^-(aq)\rightleftharpoons CH_3Cl(aq)+OH^-(aq)

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}

Thus the equilibrium constant expression for this reaction is K_c=\frac{[CH_3Cl]\times [OH^-]}{[CH_3OH]\times [Cl^-]}

5 0
3 years ago
7. State the relationship between the concentration of ions and the boiling point for these solutions.
balu736 [363]

Answer:

  • 7. Linear relationship.

  • 8. Because ions carry charge that can move freely in solution.

  • 9. Solute: NaCl

            Solvent: H₂O

Explanation:

7. Elevation of the boling point.

The elevation of the boiling point of a solution respect to the pure solvent is a colligative property, meaning that it is proportional to the concentration of the solute particles.

        \Delta T_b=K_b\times i\times C

        T_b=T_b^0+K_b\times i\times C

Where, ΔTb is the elevation on the boiling point, Tb is the (elevated) boiling of the solution, point, Tb° is the boiling point of the pure solvent, i is the Van't Hoof factor, and C is the molal concentration (molality)

Thus the relation of the boiling point with the mola concentration (molality) is linear.

Using the table and some calculations, you can determine this relationship:

Trial  Mass of water   mass NaCl    moles NaCl     m             Tb

             kg                      g                   mol            mol/kg       ºC

  1        0.1                     8.76              0.15               1.5             101.5

 2        0.1                    17.52             0.30              3.0             103.1

 3        0.1                    20.28            0.45              4.5             104.6

 4        0.1                   35.04             0.60              6.0             106.1        

The number of moles for each trial were determinated dividing the mass in grams of the solute by the molar mass (58.44g/mol).

The molality concentrations (mole of solue per kg of solvent) were determined dividing the number of moles of each trial by the amount of solvent in kg.

As you see in the table, the column of Tb increases by an approximate constant value of 1.5ºC and the column of the molal concentration increases at a constant value of 0.15mol. Thus there is a constante rate of change of 1.5ºC/0.15mol = 10ºC/mol.

Since a constant rate of change means constant slope, this you conclude that the relationship is linear.

8. Ability to conduct an electric current.

Electric current is the flow of charges.

Ions are charged particles. There are positive ions (cations) and negative ions (ions) dissolved; this is, they are free to move in solution.

Current is the rate of flow of the charge, thus the greater the number of ions in solution, the greater the flow of charge, which is a greater current.

That explains why the greater the concentration of the ions in the solution the greater the abiity fo the solution to conduct electricity.

Pure water has a low concentration of ions (H⁺ and OH⁻) thus it has a very low ability to conduct electricity. But adding a solute, like an acid, a base, or a salt (like NaCl in this case), increases the number of ions in solution, and thus a solution with a high concetration of solute will have a great ability to conduct electricity.

9. Identify the solute and solvent.

  • The<em> solute </em>is the component of a solutions that is present in relatively small quantities and is dissolved. In this case the solute is the salt, sodium chloride, NaCl.

  • The <em>solvent</em> is the component of the solution that is present in greater amount and is able to dissolve the solute. In this case the solvent is water, H₂O.

The concentrations of the solute normally are relatively low, because they are the component with less particles in the solution.

5 0
3 years ago
Properties of gases are dependent upon the pressure, temperature, and volume of the container they are in, as well as the number
OLga [1]

Answer:

False

Explanation:

Those would affect physical properties not chemical properties.

5 0
3 years ago
A chemist must prepare 800.0mL of sodium hydroxide solution with a pH of 12.10 at 25°C. She will do this in three steps: Fill a
AnnZ [28]

Answer:

0.42 g

Explanation:

<u>We have: </u>

pH = 12.10 (25 °C)

V = 800.0 mL = 0.800 L    

To find the mass of sodium hydroxide (NaOH) we can use the pH:

14 = pH + pOH

pOH = 14 - pH = 14 - 12.10 = 1.90

pOH = -log ([OH^{-}])

[OH]^{-} = 10^{-pOH} = 10^{-1.90} = 0.013 M

Now, we can find the number of moles (η) of OH:

\eta = ([OH]^{-})*V = 0.013 mol/L * 0.800 L = 1.04 \cdot 10^{-2} moles

Since we have 1 mol of OH in 1 mol of NaOH, the number of moles of NaOH is equal to 1.04x10⁻² moles.

Finally, with the number of moles we can find the mass of NaOH:

m = \eta * M

<em>Where M is the molar mass of NaOH = 39.9 g/mol </em>

m = 1.04 \cdot 10^{-2} moles * 39.9 g/mol = 0.42 g

Therefore, the mass of sodium hydroxide that the chemist must weigh out in the second step is 0.42 g.

I hope it helps you!

3 0
3 years ago
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