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500mL of 10^(-5)MNaOH is mixed with 500m...

`500mL` of `10^(-5)MNaOH` is mixed with `500mL` of `2.5xx10^(-5)M` of `Ba(OH_(2))`,To the resulting solution ,`99L` water is added ,calculate `pH` of final solution.Take `log0.303=-0.52`.

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To solve the problem, we need to calculate the pH of the final solution after mixing two bases, NaOH and Ba(OH)₂, and then diluting with water. Let's break down the solution step by step. ### Step 1: Calculate the number of millimoles of OH⁻ from NaOH Given: - Volume of NaOH = 500 mL = 0.5 L - Concentration of NaOH = \(10^{-5} \, M\) ...
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