Question

ln is an acidic indicator $(K_{Ind}={10}^{-7})$ which dissociates into aqueous acidic solution of 30 mL of 0.05 M $H_{3}P{O}_4(K_1={10}^{-3},K_2={10}^{-7},K_3={10}^{-13})$. If this solution is treated with 30 mL of NaOH solution, then what molarity of NaOH is needed to reach the equivalence point with indicator?

• A. 0.1 M
• B. 0.2 M
• C. 0.3 M
• D. 0.4 M

Answer 1

The correct option is A 0.1 M

Hln is an acidic indicator $(K_{Ind}={10}^{-7})$

30 mL of 0.05 M $H_{3}P{O}_4(K_1={10}^{-3},K_2={10}^{-7},K_3={10}^{-13})$

From acid dissociation constants it can be seen that indicator will indicate at second dissociation point so hydrogen ion conc at this point is,

$V\times M\times n_f=30\times 0.05\times 2$

At equivalence point conc. of NaOH will be same so

$30\times 0.05\times 2=30\times xM$

$x=0.1M$

Thus, the molarity of NaOH = 0.1 M

Hence, the correct option is $\text{A}$