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×[tex]\frac{1mol}{116,072g}[/tex]=7,952x10⁻³moles of H₂M
60,0mL of 0,265M KOH are:
0,0600L×[tex]\frac{0,265mol}{1L}[/tex]=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!
