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This is a titration question: we want to have the same number of hydroxide ions as hydroxide ions so that they will form water and the pH will be neutral. In chemistry, we count atoms and molecules in moles, and we can calculate how many moles of HBr we have, because concentration in molarity is the number of moles divided by the volume in liters... M = moles/V. We plug in what we got: 1.45M = moles/0.0350L, and solve for moles: 0.0508 moles.

Now we know we need 0.0508 moles of NaOH, whose molecular weight is 40g/mole. MW x moles = grams, so (40g/mole)(0.0508 moles) = 2.03 g of NaOH.

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Q: How many grams of solid sodium hydroxide would need to be added to completely neutralize 35.0 mL of 1.45 M HBr?
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