KH₂PO₄ hydrolyzes as;
H₂PO₄⁻ + H₂O ↔ H₃PO₄ + OH⁻
Let x amount of H₂PO₄⁻ has reacted with water then,
Kb₁ = [H₃PO₄][OH⁻] / [H₂PO₄⁻]
[H₂PO₄⁻] = 0.8-x M
Kb₁ = x² / (0.8 - x)
Given Ka₁ = 7.5 x 10⁻³
so Kb₁ = 1 x 10⁻¹⁴ / (7.5 x 10⁻³) = 1.33 x 10⁻¹²
From this information:
1.33 x 10⁻¹² = x² / 0.8
x = [OH⁻] = 1.03 x 10⁻⁶ M
pOH = - log (1.03 x 10⁻⁶) = 5.99
pH = 14 - pOH = 14 - 5.99 = 8.01
Answer:
HCO₂/H₂O is not the acid-base conjugate pair.
Explanation:
<em>Acid and conjugate base pairs differ by an H+ ion.</em>
Neither HCO₂ nor H₂O has lost or gained protons.
The conjugate acid of H₂O is H₃O⁺
The conjugate base of HCO₃⁻ is CO₃²⁻
[A conjugate acid has one more H⁺ than its base]
One: looks to be correct for both answers. Certainly the first one is. The second depends on your other choices. But military use is one.
Two: is correct. Pd has (in this case) an atomic mass of 114 and its number is 46
Three: Even with my slop numbers, 4.98 is the answer (although I get 4.99 but again, my numbers are pretty sloppy).
Four: Slop numbers say 78.3, but 78 is the right answer.
Five: Slop numbers agree with Al2S3. I think that's D
They are all correct. Very Fine Work.
Answer:
3 NH4OH (l) + H3PO4 (aq) → (NH4)3PO4 (aq) + 3 H2O (l)
Explanation:
This is an acid-base reaction (neutralization): NH4OH is a base, H3PO4 is an acid
