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ChemistryHelper2024
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 



mahima6824soni

The pH of the solution is  8.01.

What is pH?

pH is measuring scale of acidity and basicity of any solution.

The hydrolyzing reaction for KH₂PO₄ is

[tex]\rm H_2PO_4^- + H_2O = H_3PO_4 + OH^-[/tex]

Let x amount of H₂PO₄⁻ has reacted with water then,

[tex]Kb_1 = \dfrac{ [H_3PO_4][OH^-] }{[H_2PO_4^-]}[/tex]

[tex]\rm [H_2PO_4^-] = 0.8-x M\\\\Kb_1 =\dfrac{x^2}{ (0.8 - x)} \\\\Given, that Ka_1 = 7.5 \times 10^-^3\\\\\\then, Kb_1 = \dfrac{10^-^1^4 }{7.5 \times 10^-^3} = 1.33 \times 10^-^1^2[/tex]

[tex]\rm 1.33 \times 10^-^1^2 = \dfrac{x^2}{0.8}\\\\x = [OH^-] = 1.33 \times 10^-^6 M\\\\pOH = - log( 1.33 \times 10^-^6) = 5.99\\\\\\Now, pH = 14 - pOH \\\\14 - 5.99 = 8.01[/tex]

Thus, the pH is 8.01

Learn more about pH

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