0.15 mol of aspartic acid hydrochloride, $HOOC - CH X_{2} - CH (NH X_{3} X^{+} Cl -) - COOH$ , is treated with just enough NaOH to convert it completely into $NaOOC - CH X_{2} - CH (NH X_{2}) - COONa$ (plus NaCl and water). The amount of NaOH required is

Answer: D

Each mole of the salt has three acidic protons: two

- COOH

and one

- NH X_{3} X^{+}

(the chloride is the counter ion of the ammonium group). Moles of NaOH =

3 \times 0.15 = 0.45

mol.