Calculate the normality of each of the following solutions.

For a, we have \(4.6 \times 10^{-2}\) molar of sodium hydroxide. Recall that normality is simply \(n \times \)molarity, where \(n\) is the number of \(OH^-\) ions for the base. Since there's only one \(OH^-\) in sodium hydroxide, it will be \(1 \times 4.6 \times 10^{-2}\). Therefore, the normality will be \(4.6 \times 10^{-2}\).

For b, a bit more calculation is required. Here, we have 0.35 grams of phosphoric acid in 1 liter. We use the equation: N = e q u i v a l e n c e l i t e r s of solution. With 1 liter on the bottom, we need to determine the equivalents of this base. Recall that for a base, the equivalents are \(n \times\) moles. \(H_3PO_4\) has 3 \(H^+\) ions, hence \(n = 3\). We now need the moles of phosphoric acid, where 0.35 grams are provided. Given a molar mass of 97.994 for phosphoric acid (3 hydrogens, 1 phosphorus, and 4 oxygens), we calculate the moles which is approximately \(0.0035716\) moles. This gives \(0.0107148\) equivalents when multiplied by 3. Dividing by 1 liter results in a normality of approximately \(0.011\), considering significant figures based on the mass given (0.35 has 2 significant figures).

Thus, the answers for parts a and b are \(4.6 \times 10^{-2}\) and \(0.011\) normality respectively.