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Elena L [17]
3 years ago
9

If you have access to stock solutions of 1.00 M H3PO4, 1.00 M of HCl, and 1.00 M NaOH solution, (and distilled water of course),

what volumes of each would you mix before diluting to a final volume of 2.00 L to prepare 2.00 L of pH 7.40 buffer with a final total concentration of 50 mM of phosphorous contains species (e.g. so that [H3PO4] [H2PO4 - ] [HPO4 2- ] [PO4 3- ]
Chemistry
1 answer:
garri49 [273]3 years ago
4 0

Answer:

0.10L of 1.00M of H₃PO₄ and 0.1613L of 1.00M NaOH

Explanation:

The pKa's of phosphoric acid are:

H₃PO₄/H₂PO₄⁻ = 2.1

H₂PO₄⁻/HPO₄²⁻ = 7.2

HPO₄²⁻/PO₄³⁻ = 12.0

To make a buffer with pH 9.40 we need to convert all H₃PO₄ to H₂PO₄⁻ and an amount of H₂PO₄⁻ to HPO₄²⁻

To have a 50mM solution of phosphoures we need:

2L * (0.050mol / L) = 0.10 moles of H₃PO₄

0.10 mol * (1L / mol) = 0.10L of 1.00M of H3PO4

To convert the H₃PO₄ to H₂PO₄⁻ and to HPO₄²⁻ must be added NaOH, thus:

H₃PO₄ + NaOH → H₂PO₄⁻ + H₂O + Na⁺

H₂PO₄⁻ + NaOH → HPO₄²⁻ + H₂O + Na⁺

Using H-H equation we can find the amount of NaOH added:

pH = pKa + log [A⁻] / [HA] <em>(1)</em>

<em>Where [A-] is conjugate base, HPO₄²⁻ and [HA] is weak acid, H₂PO₄⁻</em>

<em>pH = 7.40</em>

<em>pKa = 7.20</em>

[A-] + [HA] = 0.10moles <em>(2)</em>

Replacing (2) in (1):

7.40 = 7.20 + log 0.10mol - [HA] / [HA]

0.2 = log 0.10mol - [HA] / [HA]

1.5849 = 0.10mol - [HA] / [HA]

1.5849 [HA] = 0.10mol - [HA]

2.5849[HA] = 0.10mol

[HA] = 0.0387 moles = H₂PO₄⁻ moles

That means moles of HPO₄²⁻ are 0.10mol - 0.0387moles = 0.0613 moles

The moles of NaOH needed to convert all H₃PO₄ in H₂PO₄⁻ are 0.10 moles

And moles needed to obtain 0.0613 moles of HPO₄²⁻ are 0.0613 moles

Total moles of NaOH are 0.1613moles * (1L / 1mol) = 0.1613L of 1.00M NaOH

Then, you need to dilute both solutions to 2.00L with distilled water.

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Atomic mass of 1 mole of hydrogen atom = 1 g

Mass of hydrogen in 1 mole of C_{12}H_{10}O_{16} = 10 × 1 g = 10 g

d)In 1 mole of CH_3CH_2CH_2CH_2CH_3 there are 12 moles of hydrogen atom.

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3 0
3 years ago
Find the volume of a gas at standard pressure if its volume at 1.9 atm is 80 ml?
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Answer:

1.5 × 10² mL

Explanation:

Step 1: Given data

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Step 2: Calculate the final volume of the gas

For an ideal gas, we can calculate the final volume of the gas using Boyle's law.

P₁ × V₁ = P₂ × V₂

V₂ = P₁ × V₁/P₂

V₂ = 1.9 atm × 80 mL/1.0 atm

V₂ = 1.5 × 10² mL

Since the pressure decreased, the volume of the gas increased.

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Electrical hazards may be defined as a place or a condition which is dangerous and where a person working can come in contact with electrical contact with some energized equipment or any electrical conductor. There is always danger or a chance for the workmen to get injury or get a flash burn while working.

The safety steps are :

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Answer:

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