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ohaa [14]
3 years ago
14

Identify each definition that applies to the compound in red. Check all that apply.

Chemistry
2 answers:
Tema [17]3 years ago
7 0

Answer: Bronsted Lowry base

Explanation:

Lina20 [59]3 years ago
7 0

Answer:

Bronsted-Lowry base ONLY

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Answer:

  • Mass of NaH₂PO₄·H₂O = 8.542 g
  • Mass of Na₂HPO₄ = 5.410 g

Explanation:

Keeping in mind the equilibrium:

H₂PO₄⁻ ↔ HPO₄⁻² + H⁺

We use the Henderson-Hasselbalch equation (H-H):

pH = pka + log\frac{[A^{-}]}{[HA]}

For this problem [A⁻] = [HPO₄⁻²] and [HA] = [H₂PO₄⁻]

From literature we know that pka = 7.21, from the problem we know that pH=7.00 and that

[HPO₄⁻²] + [H₂PO₄⁻] = 0.100 M

From this equation we can <u>express [H₂PO₄⁻] in terms of [HPO₄⁻²]</u>:

[H₂PO₄⁻] = 0.100 M - [HPO₄⁻²]

And then replace [H₂PO₄⁻] in the H-H equation, <u>in order to calculate [HPO₄⁻²]</u>:

7.00=7.21+log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]} \\-0.21=log\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\10^{-0.21} =\frac{[HPO4^{-2}] }{0.100 M-[HPO4^{-2}]}\\0.616*(0.100M-[HPO4^{-2}])=[HPO4^{-2}]\\0.0616 M = 1.616*[HPO4^{-2}]\\0.03812 M =[HPO4^{-2}]

With the value of  [H₂PO₄⁻],<u> we calculate [HPO₄⁻²]</u>:

[HPO₄⁻²] + 0.0381 M = 0.100 M

[HPO₄⁻²] = 0.0619 M

Finally, using the concentrations, the volume, and the molecular weights; we can calculate the weight of each substance:

  • Mass of NaH₂PO₄·H₂O = 0.0619 M * 1 L * 138 g/mol = 8.542 g
  • Mass of Na₂HPO₄ = 0.0381 M * 1 L * 142 g/mol = 5.410 g

8 0
4 years ago
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