Question

Calculate the [OH] of $[N{H}_2{C}_2{H}_{4}N{H}_3{]}^{\oplus }$ and $[H_{3}N-C_2{H}_{4}N{H}_3{]}^{2+}$ in 0.15 M ethylene diamine (aq) if
$N{H}_2{C}_2{H}_{4}N{H}_2+H_{2}O\rightleftharpoons N{H}_2{C}_2{H}_4\stackrel{\oplus }{N}{H}_3+\stackrel{⊝}{O}H(K_1=8.5\times {10}^{-5})$
$N{H}_2{C}_2{H}_4\stackrel{\oplus }{N}{H}_3+H_{2}O\rightleftharpoons [N{H}_3{C}_2{H}_{4}N{H}_3{]}^{2+}+\stackrel{⊝}{O}H(K_2=2.7\times {10}^{-8})$

• A. Case I : $3.57\times {10}^{-3}M$ Case II : $2.7\times {10}^{-8}M=K_{b_2}$
• B. Case I : $4.57\times {10}^{-3}M$ Case II : $3.7\times {10}^{-8}M=K_{b_2}$
• C. Case I : $5.57\times {10}^{-3}M$ Case II : $4.7\times {10}^{-8}M=K_{b_2}$
• D. Case I : $6.57\times {10}^{-3}M$ Case II : $5.7\times {10}^{-8}M=K_{b_2}$

Answer 1

Answer: (A)

Case I : $3.57\times {10}^{-3}M$ Case II : $2.7\times {10}^{-8}M=K_{b_2}$

Case I : $\left[O{H}^{⊝}\right]=[N{H}_2{C}_2{H}_{4}N{H}_3{]}^{\oplus }$
$=C\times \sqrt{\frac{K_{b_1}}{C}}=\sqrt{0.15\times 8.5\times {10}^{-10}}=3.57\times {10}^{-3}M$

Case II :
$N{H}_2{C}_2{H}_{4}N{H}_3^{\oplus }+H_{2}O\rightleftharpoons N{H}_3{C}_2{H}_{4}N{H}_4{]}^{2+}+O{H}^{⊝}$
$3.57\times {10}^{-3}01.275\times {10}^{-5}$

$[3.57\times {10}^{-3}-X]X(X+1.275\times {10}^{-5})$

$2.7\times {10}^{-8}=\frac{x\cdot (X+3.5\times {10}^{-3})}{(3.57\times {10}^{-3}-X)}$

Neglecting $X^2$, also $3.57\times {10}^{-3}-X=3.57\times {10}^{-3}$,
(X is very small)

$2.7\times {10}^{-8}=\frac{3.57\times {10}^{-3}X}{3.57\times {10}^{-3}}$

$\therefore X=2.7\times {10}^{-8}$

$\therefore [N{H}_2{C}_2{H}_{4}N{H}_3{]}^{2+}=2.7\times {10}^{-8}M=K_{b_2}$