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ValentinkaMS [17]
3 years ago
7

A 1 liter solution contains 0.443 M acetic acid and 0.332 M sodium acetate. Addition of 0.365 moles of perchloric acid will: (As

sume that the volume does not change upon the addition of perchloric acid.) Raise the pH slightly Lower the pH slightly Raise the pH by several units Lower the pH by several units Not change the pH Exceed the buffer capacity
Chemistry
1 answer:
Yuki888 [10]3 years ago
4 0

Answer:

Lower the pH several units

Explanation:

Typically, a buffer system would minimize changes in pH values with additions of strong acid or base. However, one should realize buffer systems do have limits. The limits are dictated by the amount of common ion in the buffer solution mix. The chemistry of the buffer and common ion functions to remove the excess strong acid or base added. However, if more strong acid or strong base is added that exceeds the capacity of the buffer solution to remove the an excess acid or base solution would result and pH values would change drastically.

In this problem, perchloric acid (HClO₄) is a strong acid delivering hydronium ion in an amount equal to the amount of perchloric acid noted in the problem.

Since the concentration of perchloric acid (HClO₄) and H⁺ ion exceeds the concentration of available acetate ion in the buffer solution, the buffer effect would be exhausted before removing all of the HClO₄. The remaining HClO₄ would then function to drastically reduce the pH of the remaining solution several units before stabilizing.

The 0.443M HOAc/0.332M NaOAc buffer before addition of the HClO₄ has a pH of 4.62.  Adding 0.365M in HClO₄ would cause the HOAc equilibrium to shift left because of the H⁺ overload. This shift would continue until all of the strong acid was removed or all of the conjugate base is exhausted by converting back to HOAc. That is, adding 0.365M HClO₄ removes all 0.332M OAc⁻ leaving 0.365M - 0.332M = 0.033M excess HClO₄. The pH of a 0.033M HClO₄ solution => pH = -logs[H⁺] = -log(0.033) = 1.48.

=> compare pH before adding acid = 4.62 to pH after adding acid = 1.48.

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The most common source of copper (cu) is the mineral chalcopyrite (cufes2). how many kilograms of chalcopyrite must be mined to
tigry1 [53]

Answer : 0.8663 Kg of chalcopyrite must be mined to obtained 300 g of pure Cu.

Solution : Given,

Mass of Cu = 300 g

Molar mass of Cu = 63.546 g/mole

Molar mass of CuFeS_2 = 183.511 g/mole

  • First we have to calculate the moles of Cu.

\text{ Moles of Cu}=\frac{\text{ Given mass of Cu}}{\text{ Molar mass of Cu}}= \frac{300g}{63.546g/mole}=4.7209moles

The moles of Cu = 4.7209 moles

From the given chemical formula, CuFeS_2 we conclude that the each mole of compound contain one mole of Cu.

So, The moles of Cu = Moles of CuFeS_2 = 4.4209 moles

  • Now we have to calculate the mass of CuFeS_2.

Mass of CuFeS_2 = Moles of CuFeS_2 × Molar mass of CuFeS_2 = 4.4209 moles × 183.511 g/mole = 866.337 g

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3 0
3 years ago
Which solution contains the smallest number of moles of sucrose (c12h22o11, molar mass = 342.30 g/mol)? 2,000 ml of a 5.0 × 10–5
Iteru [2.4K]

> 2,000 mL of a 5.0 × 10–5% (w/v) sucrose solution 

5.0 × 10–3 g/mL * 2000 mL * (1 mol / 342.30 g) = 0.0292 mol

<span>
> 2,000 mL of a 5.0 ppm sucrose solution</span>

5 grams / 1000000 mL * 2000 mL* (1 mol / 342.30 g) = 0.0000292 mol

 <span>
> 20 mL of a 5.0 M sucrose solution </span>

5.0 M * 0.020 L = 0.1 mol

 

 

Answer:

<span>2,000 mL of a 5.0 ppm sucrose solution</span>

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3 years ago
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