1. Asam ftalat H_(2)C_(8)H_(4)O_(4) adalah asam diprotic yang digunakan pada pembuatan indicator fenolftalein. K_(a1)=1,2times 10^-13, dan Ka2=3,9times 10^-6 hitung: a. Konsentrasi ion H^+ darei larutan fenolftalein 0,015 m! b. Konsentrasi ion ftalat, C_(8)H_(4)O_(4)^2- dalam larutan tersebut!

The question asks to calculate the concentration of H⁺ ions and the phthalate ion (C₈H₄O₄²⁻) in a 0.015 M phthalic acid solution, given the Ka1 and Ka2 values. Since Ka1 is much smaller than Ka2 (1.2 x 10⁻¹³ << 3.9 x 10⁻⁶), we can simplify the calculation by considering only the second ionization step. The first ionization contributes negligibly to the overall H⁺ concentration.**a. Concentration of H⁺ ions:**The second ionization of phthalic acid is represented as:HC₈H₄O₄⁻ ⇌ H⁺ + C₈H₄O₄²⁻We can use the Ka2 expression:Ka2 = [H⁺][C₈H₄O₄²⁻] / [HC₈H₄O₄⁻]Let x be the concentration of H⁺ ions produced in the second ionization. Since the stoichiometry is 1:1, the concentration of C₈H₄O₄²⁻ will also be x, and the concentration of HC₈H₄O₄⁻ will be approximately 0.015 - x. Because Ka2 is small, we can assume that x is much smaller than 0.015, so we can simplify the equation to:Ka2 ≈ x² / 0.015Solving for x:x = √(Ka2 * 0.015) = √(3.9 x 10⁻⁶ * 0.015) = √(5.85 x 10⁻⁸) ≈ 2.42 x 10⁻⁴ MTherefore, the concentration of H⁺ ions is approximately **2.42 x 10⁻⁴ M**.**b. Concentration of C₈H₄O₄²⁻ ions:**As we assumed in part (a), the concentration of C₈H₄O₄²⁻ ions is equal to x. Therefore, the concentration of C₈H₄O₄²⁻ ions is approximately **2.42 x 10⁻⁴ M**.**Important Note:** This calculation simplifies the problem by neglecting the contribution of the first ionization. A more rigorous calculation would involve solving a quadratic equation considering both ionization steps, but given the significant difference in Ka1 and Ka2 values, the simplification provides a reasonably accurate approximation.