Calculate the pH of a solution formed when 1.5 L oF₄.5×10^{- 3} M HCl is mixed witH₂ .0 L of 3.0×10^{- 3} M HNO₃.

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To calculate the pH of the solution, first, we need to find the total moles of H+ ions contributed by both HCl and HNO3, as both are strong acids and dissociate completely in water.For HCl: $1.5 \, \text{L} \times 0.5 \times 10^{-3} \, \text{M} = 7.5 \times 10^{-4} \, \text{moles}$For HNO3: $2.0 \, \text{L} \times 3.0 \times 10^{-3} \, \text{M} = 6.0 \times 10^{-3} \, \text{moles}$Total moles of H+ ions = $7.5 \times 10^{-4} + 6.0 \times 10^{-3} = 6.75 \times 10^{-3} \, \text{moles}$Total volume of the solution = $1.5 \, \text{L} + 2.0 \, \text{L} = 3.5 \, \text{L}$Concentration of H+ ions in the solution = $\frac{6.75 \times 10^{-3} \, \text{moles}}{3.5 \, \text{L}} = 1.93 \times 10^{-3} \, \text{M}$pH = -log[H+] = -log(1.93 × 10^-3) ≈ 2.71However, none of the provided options exactly matches the calculated pH of 2.71. Given the choices, it seems there might have been a miscalculation or misunderstanding in the question's details or in the calculation process. Based on the closest match principle and assuming a potential rounding or calculation difference not accounted for here, option E (2.44) would be the closest to the calculated value, but it's important to note the discrepancy in the calculation process.

Calculate the PH of a Solution Formed When 1