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expeople1 [14]
3 years ago
11

Sodium benzoate, the sodium salt of benzoic acid, is used as a food preservative. A sample containing solid sodium benzoate mixe

d with sodium chloride is dissolved in 50.0 mL of .5 M HCl, giving an acidic solution (benzoic acid mixed with HCl ). This mixture is then titrated with .393 M NaOH. After the addition of 46.50 mL of the solution, the pH is found to be 8.2. At this point, the addition of one more drop ( .02 mL) of NaOH raises the pH to 9.3 . Calculate the mass of sodium benzoate (NaC6H5COO) in the original sample. (Hint: At the equivalence point, the total number of moles of acid [here HCl] equals the total number of moles of base [here, both NaOH and NaC6H5COO]
Chemistry
1 answer:
GalinKa [24]3 years ago
3 0

Answer:

Protegerte Freddy Gerrard Reverte Freddy Buffett irrefutables

Explanation:

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A concentrated binary solution containing mostly species 2 (but x2 ≠ 1) is in equilib- rium with a vapor phase containing both s
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Answer:

x1= 4.5 × 10^-3, y1= 0.9

Explanation:

A binary solution having two species is in equilibrium in a vapor phase comprising of species 1 and 2

Take the basis as the pressure of the 2 phase system is 1 bar. The assumption are as follows:

1. The vapor phase is ideal at pressure of 1 bar

2. Henry's law apply to dilute solution only.

3. Raoult's law apply to concentrated solution only.

Where,

Henry's constant for species 1 H= 200bar

Saturation vapor pressure of species 2, P2sat= 0.10bar

Temperature = 25°C= 298.15k

Apply Henry's law for species 1

y1P= H1x1...... equation 1

y1= mole fraction of species 1 in vapor phase.

P= Total pressure of the system

x1= mole fraction of species 1 in liquid phase.

Apply Raoult's law for species 2

y2P= P2satx2...... equation 2

From the 2 equations above

P=H1x1 + P2satx2

200bar= H1

0.10= P2sat

1 bar= P

Hence,

P=H1x1 + (1 - x1) P2sat

1bar= 200bar × x1 + (1 - x1) 0.10bar

x1= 4.5 × 10^-3

The mole fraction of species 1 in liquid phase is 4.5 × 10^-3

To get y, substitute x1=4.5 × 10^-3 in equation 1

y × 1 bar = 200bar × 4.5 × 10^-3

y1= 0.9

The mole fraction of species 1 in vapor phase is 0.9

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Answer:

588.2 mL

Explanation:

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First we <u>calculate how many Fe⁺² moles reacted</u>, using the given <em>concentration and volume of FeSO₄ solution</em> (the number of FeSO₄ moles is equal to the number of Fe⁺² moles):

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Then we convert Fe⁺² moles to KOH moles, using the stoichiometric ratios:

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