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A solution of a weak acid was titrate...

A solution of a weak acid was titrated was NaOH. The equivalence point was reached when `36.12` ml of `0.10` (N) NaOH have been added . Now `18.06` ml of `0.10` (N) HCl was addd to the titrated solution . The pH was found to be `4.92 ` .What is `K_(a)` of the acid ?

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To find the \( K_a \) of the weak acid, we will follow these steps: ### Step 1: Calculate the moles of NaOH used Given: - Volume of NaOH = 36.12 mL = 0.03612 L - Normality of NaOH = 0.10 N Using the formula for moles: ...
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