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Svetach [21]
3 years ago
8

(30 Points) You need to produce a buffer solution that has a pH of 5.02. You already have a solution that contains 10. mmol (mil

limoles) of acetic acid. How many millimoles of acetate (the conjugate base of acetic acid) will you need to add to this solution? The pKa of acetic acid is 4.74. Express your answer numerically in millimoles.
Chemistry
2 answers:
chubhunter [2.5K]3 years ago
8 0
<span>By definition:

pH = pKa + log [acetate]/ [acetic acid]

so

5.02 = 4.74 + log [acetate] / 10 mmole

10mmole = 10/1000 = 0.01 mole

5.02 = 4.74 + log [acetate] / 0.01

5.02 - 4.74 = 0.28 = log [acetate] /0.01


10^0.28 = </span><span>1.90546</span> = [acetate] / 0.01 <span>
[acetate] = 0.019 mole 

= 19 millimoles

</span>
ElenaW [278]3 years ago
3 0
<span>pH = pKa + log [acetate]/ [acetic acid] ( Handerso-Hasselbalch equation)

5.35 = 4.74 + log acetate / 0.010 moles

5.35 - 4.74 =0.61 = log acetate / 0.010

10^0.61 =4.1 = acetate / 0.010

moles acetate = 0.041

millimoles acetate required = 41</span>
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Fluorine. 

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25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base soluti
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Answer:

[HF]₀ = 0.125M

Explanation:

NaOH + HF => NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3.   This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

                HF  ⇄    H⁺    +      F⁻

C(eq)       [HF]     10⁻³M      0.089M (<= soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka

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3 years ago
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Answer: 72 grams of O_2(g) are needed to completely burn 19.7 g C_3H_8(g)

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To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Putting in the values we get:

\text{Number of moles}=\frac{19.7g}{44g/mol}=0.45moles

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According to stoichiometry:

1 mole of C_3H_8 requires 5 moles of oxygen

0.45 moles of C_3H_8 require= \frac{5}{1}\times 0.45=2.25 moles of oxygen

Mass of O_2=moles\times {\text {Molar mass}}=2.25\times 32=72g

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Option C, Antarctic Treaty System, is the right answer.

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algol13

Answer:

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