Match the equivalent weight of H 3 P O 4 (molar mass 98) in Column I with the chemical change in Column II. Column I Column II A) 98 P) H 3 P O 4 + NaOH → Na H 2 P O 4 B) 49 Q) H 3 P O 4

Question

Match the equivalent weight of H 3 ​ P O 4 ​ (molar mass 98) in Column I with the chemical change in Column II. Column I Column II A) 98 P) H 3 ​ P O 4 ​ + NaOH → Na H 2 ​ P O 4 ​ B) 49 Q) H 3 ​ P O 4 ​ + 2NaOH → N a 2 ​ HP O 4 ​ C) 32.67 R) H 3 ​ P O 4 ​ + 3NaOH → N a 3 ​ P O 4 ​

TestHub · Accepted Answer

Equivalent weight = 98 / n where n is the number of H + replaced. (P) Only one H + neutralized (to Na H 2 ​ P O 4 ​ ): n = 1, E = 98 (A). (Q) Two H + neutralized (to N a 2 ​ HP O 4 ​ ): n = 2, E = 49 (B). (R) All three H + neutralized (to N a 3 ​ P O 4 ​ ): n = 3, E = 98/3 = 32.67 (C).

Column I	Column II
A) 98	P) $H_{3} P O_{4}$ + $NaOH$ $\to$ $Na H_{2} P O_{4}$
B) 49	Q) $H_{3} P O_{4}$ + 2NaOH $\to$ $N a_{2} HP O_{4}$
C) 32.67	R) $H_{3} P O_{4}$ + 3NaOH $\to$ $N a_{3} P O_{4}$

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