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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 +
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>:
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