A 100.0 mL solution containing 0.877 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.256 M KOH. Calculate the pH of the solution after the addition of 59.0 mL of the KOH solution. Maleic acid has pKa 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 H, M, HM¯, and M2-, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. [M²-] = M %3D [HM-] = M [H,M] = M
A 100.0 mL solution containing 0.877 g of maleic acid (MW = 116.072 g/mol) is titrated with 0.256 M KOH. Calculate the pH of the solution after the addition of 59.0 mL of the KOH solution. Maleic acid has pKa 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 H, M, HM¯, and M2-, which represent the fully protonated, intermediate, and fully deprotonated forms, respectively. [M²-] = M %3D [HM-] = M [H,M] = M
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![**Titration of Maleic Acid with KOH**
A 100.0 mL solution containing 0.877 g of maleic acid (Molecular Weight = 116.072 g/mol) is titrated with 0.256 M KOH. Calculate the pH of the solution after the addition of 59.0 mL of the KOH solution. Maleic acid has \( pK_a \) values of 1.92 and 6.27.
**pH Calculation:**
pH = [Input box for pH]
**Equilibrium Concentrations of Maleic Acid Forms**
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 \( \text{H}_2\text{M}, \text{HM}^-, \) and \(\text{M}^{2-}\), which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.
\[
[\text{M}^{2-}] = \text{[Input box]} \, \text{M}
\]
\[
[\text{HM}^-] = \text{[Input box]} \, \text{M}
\]
\[
[\text{H}_2\text{M}] = \text{[Input box]} \, \text{M}
\]
**Instructions**
1. Complete the calculations for pH after the titration.
2. Calculate the equilibrium concentrations of each form of maleic acid using the provided \( pK_a \) values and the calculated pH.
This exercise enables students to understand the relationship between acid-base titrations, pH, and the equilibrium concentrations of the different forms of a diprotic acid.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5407ea30-1dc6-4393-bdac-fa3faec37927%2F4e134dd5-96f9-468b-a192-55af4cc0942e%2Fqjh62ti_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Titration of Maleic Acid with KOH**
A 100.0 mL solution containing 0.877 g of maleic acid (Molecular Weight = 116.072 g/mol) is titrated with 0.256 M KOH. Calculate the pH of the solution after the addition of 59.0 mL of the KOH solution. Maleic acid has \( pK_a \) values of 1.92 and 6.27.
**pH Calculation:**
pH = [Input box for pH]
**Equilibrium Concentrations of Maleic Acid Forms**
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 \( \text{H}_2\text{M}, \text{HM}^-, \) and \(\text{M}^{2-}\), which represent the fully protonated, intermediate, and fully deprotonated forms, respectively.
\[
[\text{M}^{2-}] = \text{[Input box]} \, \text{M}
\]
\[
[\text{HM}^-] = \text{[Input box]} \, \text{M}
\]
\[
[\text{H}_2\text{M}] = \text{[Input box]} \, \text{M}
\]
**Instructions**
1. Complete the calculations for pH after the titration.
2. Calculate the equilibrium concentrations of each form of maleic acid using the provided \( pK_a \) values and the calculated pH.
This exercise enables students to understand the relationship between acid-base titrations, pH, and the equilibrium concentrations of the different forms of a diprotic acid.
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