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lara [203]
3 years ago
14

What is the pH of a solution of 0.800 M KH2PO4, potassium dihydrogen phosphate?

Chemistry
1 answer:
Nookie1986 [14]3 years ago
3 0
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 



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Answer:

V KOH = 41 mL

Explanation:

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⇒ V KOH = ( V×<em>C </em>) acid / <em>C </em>KOH

⇒ V KOH = (0.041 L)(0.0050 mol/L) / (0.0050 mol/L)

⇒ V KOH = 0.041 L

4 0
3 years ago
Chemistry omg need help plz I need plz. I real badly at it and try to get better but plz help me
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7 0
3 years ago
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The list from reactant to final product:

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