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SashulF [63]
4 years ago
14

What is the name of H3PO4?

Chemistry
2 answers:
beks73 [17]4 years ago
7 0

Answer:

The compound is called phosphoric acid. It is also called orthophosphoric acid.

Explanation:

Hello!

Let's solve this!

The compound is:

H3PO4

It consists of three hydrogen atoms, one phosphorus atom and four oxygen atoms.

The compound is called phosphoric acid. It is also called orthophosphoric acid.

The other types of acids with phosphorus are:

metaphosphoric acid (HPO3).

pyrophosphoric acid (H4P2O7).

guajiro [1.7K]4 years ago
4 0
<span>Formula: </span><span>H3PO<span>4
Name: </span></span>Phosphoric acid
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Which solution has a higher percent ionization of the acid, a 0.10M solution of HC2H3O2(aq) or a 0.010M solution of HC2H3O2(aq)
Svetach [21]

Answer:

The solution 0.010 M has a higher percent ionization of the acid.

Explanation:

The percent ionization can be found using the following equation:

\% I = \frac{[H_{3}O^{+}]}{[CH_{3}COOH]} \times 100    

Since we know the acid concentration in the two cases, we need to find [H₃O⁺].          

By using the dissociation of acetic acid in the water we can calculate the concentration of H₃O⁺ in the two cases:

1. Case 1 (0.1 M):

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺(aq)   (1)

0.1 - x                                         x                     x

Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}  (2)

Where:

Ka: is the dissociation constant of acetic acid = 1.7x10⁻⁵.

1.7 \cdot 10^{-5} = \frac{x^{2}}{0.1 - x}  

1.7 \cdot 10^{-5}*(0.1 - x) - x^{2} = 0

By solving the above equation for x we have:

x = 1.29x10⁻³ M = [CH₃COO⁻] = [H₃O⁺]

Hence, the percent ionization is:      

\% I = \frac{1.29 \cdot 10^{-3} M}{0.1 M} \times 100 = 1.29 \%                      

   

2. Case 2 (0.01 M):

The dissociation constant from reaction (1) is:

Ka = \frac{[CH_{3}COO^{-}][H_{3}O^{+}]}{[CH_{3}COOH]}

With [CH₃COOH] = 0.01 M

1.7 \cdot 10^{-5} = \frac{x^{2}}{0.01 - x}  

1.7 \cdot 10^{-5}*(0.01 - x) - x^{2} = 0

By solving the above equation for x:

x = 4.04x10⁻⁴ M = [CH₃COO⁻] = [H₃O⁺]    

Then, the percent ionization for this case is:

\% I = \frac{4.04 \cdot 10^{-4} M}{0.01 M} \times 100 = 4.04 \%

As we can see, the solution 0.010 M has a higher percent ionization of the acetic acid.

Therefore, the solution 0.010 M has a higher percent ionization of the acid.

I hope it helps you!    

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