what is the pH of a .0020M NaOH solution?

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You need to evaluate pH using the following equation, such that:

`pH = -log(NaOH)`

Since the problem provides the concentration of `0.002M` , you need to replace it in formula above:

`pH = -log(0.002) => pH = -(-2.69) => pH = 2.69`

**Hence, evaluating the pH of the `.002 M NaOH` solution yields **`pH = 2.69.`

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