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nekit [7.7K]
3 years ago
15

Ammonium phosphate is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid with ammonia . Wha

t mass of ammonium p
Chemistry
1 answer:
Dmitry [639]3 years ago
6 0

Answer:

7.5 g

Explanation:

There is some info missing. I think this is the original question.

<em>Ammonium phosphate ((NH₄)₃PO₄) is an important ingredient in many fertilizers. It can be made by reacting phosphoric acid (H₃PO₄) with ammonia (NH₃). What mass of ammonium phosphate is produced by the reaction of 4.9 g of phosphoric acid? Be sure your answer has the correct number of significant digits.</em>

<em />

Step 1: Write the balanced equation

H₃PO₄ + 3 NH₃ ⇒ (NH₄)₃PO₄

Step 2: Calculate the moles corresponding to 4.9 g of phosphoric acid

The molar mass of phosphoric acid is 98.00 g/mol.

4.9 g \times \frac{1mol}{98.00g} = 0.050mol

Step 3: Calculate the moles of ammonium phosphate produced from 0.050 moles of phosphoric acid

The molar ratio of H₃PO₄ to (NH₄)₃PO₄ is 1:1. The moles of (NH₄)₃PO₄ produced are 1/1 × 0.050 mol = 0.050 mol.

Step 4: Calculate the mass corresponding to 0.050 moles of ammonium phosphate

The molar mass of ammonium phosphate is 149.09 g/mol.

0.050mol \times \frac{149.09 g}{mol} = 7.5 g

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We will abbreviate malonic acid CH2(CO2H)2, a diprotic acid, as H2A (pK1 = 2.847 and pK2 = 5.696). Find the pH in (a) 0.200 M H2
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Answer:

a. pH  → 1.77

b. pH → 4.27

Explanation:

Malonic acid is a dyprotic acid. It releases two protons:

H₂A  +  H₂O →  H₃O⁺   +  HA⁻     Ka1

HA⁻   +  H₂O →  H₃O⁺   +  A⁻²          Ka2

Let's find the first pH:

We expose the mass balance:

Ca = [HA]  +  [HA⁻] + [A⁻²] = 0.2 M

We can not consider the [A⁻²] so → Ca =  [HA]  +  [HA⁻] = 0.2M

As the acid is so concentrated, we can not consider the HA- so:

Ca = [HA] = 0.2 M

Charge balance →  [H⁺]  =  [HA⁻]  + [OH⁻]

H₂A  +  H₂O →  H₃O⁺   +  HA⁻     Ka1

Ka = H₃O⁺ . HA⁻ / H₂A

We need the HA⁻ value to put on the charge balance. We re order the Ka expression:

HA⁻ = Ka . H₂A / H₃O⁺           (notice that H₃O⁺ = H⁺)

We replace:  H⁺¨ = Ka . H₂A / H⁺

(H⁺)² = Ka . Ca

H⁺ = √(Ka . Ca)         We determine Ka from pKa → 10²'⁸⁴⁷ = 1.42×10⁻³

H⁺ = √(1.42×10⁻³ . 0.2)  =  0.016866

- log [H⁺] = pH  → - log 0.016866 = 1.77

b. NaHA is the salt from the weak acid, where the HA⁻ works as an amphoterous, this means that can be an acid or a base:

HA⁻  +  H₂O  ⇄  A⁻²  +  H₃O⁺     Ka₂

HA⁻  +  H₂O ⇄  H₂A  +  OH⁻      Kb2

There is a formula than can predict the pH, so now we need to compare the Ka₂ and Kb₂ data.

Ka₂ = 10⁻⁵'⁶⁹⁶ = 2.01×10⁻⁶

Kb₂ = 1×10⁻¹⁴ / 1.42×10⁻³  = 7.04×10⁻¹²

So Ka₂ > Kb₂. In conclussion the pH will be acid.

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