Question

A 100.0 mL solution containing 0.923 gof maleic acid (MW=116.072 g/mol) is titrated with 0.265 M KOH. Calculate the pH of the solution after the addition of 60.0 mL of the KOH solution. Maleic acid has pKapKa values of 1.92 and 6.27.
pH=


At this pH, calculate the concentration of each form of maleic acid in the solution at equilibrium. The three forms of maleic acid are abbreviated as H2M, HM−, and M2− which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.


[M2−]=__M


[HM−]=__M


[H2M]=__M

Answer 1

Answer:

pH = 9,57

[M²⁻] = 7,948x10⁻²M

[HM⁻] = 4x10⁻⁵M

[H₂M] = 0M

Explanation:

The moles of maleic acid presents in the solution are:

0,923g×$\frac{1mol}{116,072g}$=7,952x10⁻³moles of H₂M

60,0mL of 0,265M KOH are:

0,0600L×$\frac{0,265mol}{1L}$=0,0159 moles of KOH

The reactions of maleic acid (H₂M) and then with HM⁻ are:

H₂M + KOH → HM⁻ + H₂O + K⁺ (1)

HM⁻ + KOH → M²⁻ + H₂O + K⁺ (2)

For a complete transformation of H₂M in HM⁻ there are necessaries 7,952x10⁻³moles of KOH. As the moles of KOH are 0,0159 moles, the restant moles are:

0,0159 - 7,952x10⁻³ = 7,948x10⁻³ moles of KOH

By (2), the moles produced of M²⁻ are the same as moles of KOH, 7,948x10⁻³  moles, and moles of HM⁻ are:

7,952x10⁻³ - 7,948x10⁻³ = 4x10⁻⁶ moles of HM⁻

Using Henderson-Hasselbalch formula:

pH = pka + log₁₀ [M²⁻] /[HM⁻]

pH = 6,27 + log₁₀ 7,948x10⁻³ / 4x10⁻⁶

pH = 9,57

The moles of M²⁻ are 7,948x10⁻³  and volume of the solution is 0,1000L,

[M²⁻] = 7,948x10⁻²M

Moles of HM⁻ are 4x10⁻⁶:

[HM⁻] = 4x10⁻⁵M

And there is not H₂M:

[H₂M] = 0M

I hope it helps!