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ira [324]
2 years ago
6

Look at Table 4 in the procedure portion of the experiment. Calculate the pH you would expect each of the buffer solutions (A, B

, C, D, and E) to be using the Henderson-Hasselbalch equation, assuming that the solutions of acetic acid and sodium acetate are equimolar.
Chemistry
1 answer:
ivann1987 [24]2 years ago
4 0

The pH of the buffer solutions as determined using the Henderson–Hasselbalch equation are:

  • A. pH = 4.75
  • B. pH = 4.05
  • C. pH = 3.75
  • D. pH = 5.75
  • E. pH = 5.45

<h3>What is the pH of the solutions?</h3>

The pH of a buffer is determined using the Henderson–Hasselbalch equation shown below:

  • pH = pKₐ + log([A⁻]/[HA])

A. Volume of acetic acid = 5 mL; Volume of sodium acetate = 5 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(1)

pH = 4.75

B. Volume of acetic acid = 5 ml; Volume of sodium acetate = 1 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(1/5)

pH = 4.05

C. Volume of acetic acid = 10 ml; Volume of sodium acetate = 1 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(1/10)

pH = 3.75

D. Volume of acetic acid = 1 ml; Volume of sodium acetate = 10 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(10/1)

pH = 5.75

E. Volume of acetic acid = 1 ml; Volume of sodium acetate = 5 mL; pka of acetic acid = 4.75

The solutions of acetic acid and sodium acetate are equimolar;

pH = 4.75 + log(5/1)

pH = 5.45

In conclusion, the pH of the buffer solutions are determined using the Henderson–Hasselbalch equation.

Learn more about buffers at: brainly.com/question/22390063

#SPJ1

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german

Answer : The correct option is, (A) AlCl_3

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) AlCl_3

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of AlCl_3 will be,

AlCl_3\rightarrow Al^{3+}+3Cl^{-}

So, Van't Hoff factor = Number of solute particles = Al^{3+}+3Cl^{-} = 1 + 3 = 4

(B) KI

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of KI will be,

KI\rightarrow K^{+}+I^{-}

So, Van't Hoff factor = Number of solute particles = K^{+}+I^{-} = 1 + 1 = 2

(C) CaCl_2

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of CaCl_2 will be,

CaCl_2\rightarrow Ca^{2+}+2Cl^{-}

So, Van't Hoff factor = Number of solute particles = Ca^{2+}+2Cl^{-} = 1 + 2 = 3

(D) MgSO_4

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of MgSO_4 will be,

MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}

So, Van't Hoff factor = Number of solute particles = Mg^{2+}+SO_4^{2-} = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, AlCl_3

7 0
3 years ago
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Answer:

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Answer: 5.747 * 10^14 Hz

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If c = wavelength * frequency, where c is the speed of light (3.0 * 10^8 m/s), then you can manipulate the equation to solve for frequency (f).

f = c / wavelength

Plug in the given numbers:

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