What is the concentration of Ca²⁺ in ppm (mg/L)in cave water saturated with calcite (calcium carbonate,K_sp = 4.7 × 10^-9)?

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The equilibrium reaction for Ca²⁺ with calcite can be written as: Ca²⁺ + CaCO3 ⇌ Ca2+ + 2HCO3- + S1U1P2- Using the given Ksp value of calcite (K_sp = 4.7 × 10^-9), we can set up an expression for the equilibrium constant (Keq) for the above reaction: Keq = ([Ca2+][HCO3-]2[S1U1P2-])/([Ca²⁺][CaCO3]) = K_sp Since the concentration of CaCO3 is typically much larger than the concentration of Ca²⁺, we can assume that the concentration of CaCO3 remains constant at its saturation concentration (which depends on factors such as temperature and pressure). This allows us to simplify the above expression to: ([Ca2+][HCO3-]2[S1U1P2-])/[Ca²⁺] = K_sp/[CaCO3] We are given the value of K_sp (4.7 × 10^-9), and we can assume a typical concentration for [HCO3-] in cave water (~200 ppm or mg/L). We also know the value of [Ca²⁺] that we want to find. Substituting these values and solving for [Ca2+], we get: [Ca2+] = [HCO3-]√(K_sp/[CaCO3][S1U1P2-][Ca²⁺]) We can convert this concentration of [Ca2+] to [Ca²⁺] using the stoichiometric coefficients from the balanced equation above: [Ca²⁺] = [Ca2+]/2 Plugging in the values, we get: [Ca²⁺] = [HCO3-]√(K_sp/[CaCO3][S1U1P2-][Ca2+]) [Ca²⁺] = (200 mg/L) √(4.7 × 10^-9/(100 mg/L)(1 mg/L)([Ca2+]/2)) [Ca²⁺] = (200 mg/L) √(4.7 × 10^-9/[Ca2+]) [Ca²⁺] = (200 mg/L) √(4.7 × 10^-9/(2 × 2.5 mg/L)) [Ca²⁺] ≈ 6.9 ppm Therefore, the answer is B) 6.9 ppm.

What Is the Concentration of Ca2+ in Ppm (Mg/L)in Cave

What Is the Concentration of Ca²⁺ in Ppm (Mg/L)in Cave