0.200 mol of H3PO4 is treated with 0.520 mol NaOH so that all the acid is converted into a mixture of Na2HPO4 and Na3PO4 with no NaOH left over. The mole ratio Na3PO4:Na2HPO4 in the product is:

JEE Advanced levelSingle correct optionDifficulty 3 of 3eq146

0.200 mol of $H X_{3} PO X_{4}$ is treated with 0.520 mol NaOH so that all the acid is converted into a mixture of $Na X_{2} HPO X_{4}$ and $Na X_{3} PO X_{4}$ with no NaOH left over. The mole ratio $Na X_{3} PO X_{4} : Na X_{2} HPO X_{4}$ in the product is:

Let x = mol

Na X_{2} HPO X_{4}

and y = mol

Na X_{3} PO X_{4}

. Phosphorus balance: x + y = 0.200. Sodium (NaOH) balance: 2x + 3y = 0.520. Solving gives y = 0.120 and x = 0.080, so y : x = 3 : 2.