普通化學-酸鹼題目

Ka(H₂PO₄⁻) = 6.2X10⁻⁸


H₂PO₄⁻ <==> H⁺ + HPO₄²⁻ [ <==> is the equilibrium symbol ] Ka = [H⁺][HPO₄²⁻]/[H₂PO₄⁻]

logKa = logH + log{[HPO₄²⁻]/[H₂PO₄⁻]}

-logKa = -logH - log{[HPO₄²⁻]/[H₂PO₄⁻]}


pKa = pH - log{[HPO₄²⁻]/[H₂PO₄⁻]}


pH = pKa + log{[HPO₄²⁻]/[H₂PO₄⁻]}

pH = -log(6.2X10⁻⁸) + log{0.34/0.65)} = 7.21 - 0.28 = 6.93



increase the pH by 0.2 pH units
6.93 + 0.2 = 7.13


7.13 = 7.21 + log{[HPO₄²⁻]/[H₂PO₄⁻]}


log{[HPO₄²⁻]/[H₂PO₄⁻]} = 7.13 - 7.21 = -0.08


[HPO₄²⁻]/[H₂PO₄⁻] = 0.832


H₂PO₄⁻ <==>: H⁺ + HPO₄²⁻ [ adding x mole of N aOH to change H₂PO₄⁻ to HPO₄²⁻ ]


(0.34 + x)/(0.65 - x) = 0.832


0.34 + x = (0.832)(0.65 - x) = 0.5408 - 0.832x


1.832x = 0.5408 - 0.34 = 0.2008


x = 0.1096
Mass of N aOH added = (0.1096)(40) = 4.384 g

Ans : C