A \(0.210-g\) sample of an acid (molar mass \(=192 \mathrm{~g} / \mathrm{mol}\) ) is titrated with \(30.5 \mathrm{~mL}\) of \(0.108 \mathrm{M} \mathrm{NaOH}\) to a phenolphthalein end point. Is the acid monoprotic, diprotic, or triprotic?

We are given the concentration of NaOH solution (0.108 M) and the volume of the solution used during the titration (30.5 mL). The moles of NaOH can be calculated using the following formula: Moles of NaOH = Concentration × Volume (in Liters) Keep in mind that we need to convert the volume from mL to L by dividing the value by 1,000. Moles of NaOH = 0.108 M × (30.5 mL / 1000) = 0.003294 \(mol\)

We are given the mass of acid sample (0.210 g) and the molar mass of the acid (192 g/mol). Moles of Acid = (Mass of Acid) / (Molar Mass of Acid) Moles of Acid = 0.210 g / 192 g/mol = 0.001094 \(mol\)

Now we need to find the ratio between the moles of acid and moles of NaOH to determine the number of protons the acid donates per mole: Mole Ratio = Moles of Acid / Moles of NaOH = 0.001094 / 0.003294 Mole Ratio ≈ 0.332 Now let's compare the mole ratio value to the possible proton numbers for monoprotic, diprotic, and triprotic acids: - If the acid is monoprotic, the mole ratio would be close to 1:1 (1 proton per mole acid). - If the acid is diprotic, the mole ratio would be close to 1:2 (2 protons per mole acid). - If the acid is triprotic, the mole ratio would be close to 1:3 (3 protons per mole acid). Based on our calculated mole ratio of approximately 0.332, which is very close to 1/3, we can conclude that the given acid is: Triprotic (donates 3 protons per mole of acid).