A 0.210-g sample of an acid (molar mass = 192 g/mol)is titrated with [$X] mL of [$Y] M NaOH to a phenolphthalein endpoint.The formula of the acid is:

Question

Quizplus · Accepted Answer

We can use the formula: moles acid = moles base at equivalence point. Moles acid = 0.210 g/molar mass = 0.00109375 mol. Moles base = [$Y] M x [$X] L = [$Y][$X] mol. At equivalence point: 0.00109375 mol acid = [$Y][$X] mol NaOH. We know the molar mass of the acid is 192 g/mol. From the balanced equation: NaOH + HA → NaA + H2O, we know that the mole ratio of NaOH to acid is 1:1, which means that 0.00109375 mol of acid contains 0.00109375 mol of NaOH. Therefore: [$Y][$X] = 0.00109375 mol NaOH / 1 mol acid = 0.00109375 mol / 0.210 g/molar mass = 0.0057 M. Substituting in the values given in the question, [$Y] = 0.0057 M, which is equivalent to a pOH of 2.24. By subtracting this from 14 (pH + pOH = 14), the pH of the solution is 11.76. Phenolphthalein turns pink at a pH of around 8.2-10, indicating that the equivalence point has been reached. The only answer choice that has a potential hydroxide concentration of 0.0057 M is C, with the acid being H₃A.

A 0210-G Sample of an Acid (Molar Mass = 192 G/mol)is