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Chemistry Question on Equilibrium

2.5 ml of . 25\frac{ 2 }{ 5 } M weak monoacidic base (Kp=1×10−12 at 25∘ ( K_p = 1 \times 10^{ - 12 } \, at \, 25^\circ C) is titrated with 215\frac{ 2 }{ 1 5 } M HCI in water at 25∘ 25^\circ C. The concentration of H ' at equivalence point is (Ksp=1×10−14 at 25∘C)( K_{ sp } = 1 \times 10^{ - 14 } \, at \, 25^\circ C)

A

3.7×10−13 3.7 \times 10^{ - 13 } M

B

3.2×10−7 3.2 \times 10^{ - 7 } M

C

3.2×10−2 3.2 \times 10^{ - 2 } M

D

2.7×10−2 2.7 \times 10^{ - 2 } M

Answer

2.7×10−2 2.7 \times 10^{ - 2 } M

Explanation

Solution

mmol of base = 2.5 ×25=1\times \frac{ 2 }{ 5 } = 1
mmol of acid required to reach the end point = 1
Volume of acid required to reach the end point = 152 \frac{ 15 }{ 2}
Total volume at the end point = 152+2.5=10 \frac{ 15 }{ 2} + 2.5 = 10 ml.
Molarity of salt at the end point = 10=−.1−\frac{ 1 }{ 0} = -.1-
\begin{array} \ B^+ + \ \ \ \ \ H_2 O \rightleftharpoons BOH + \ \ \ \ \ H^ + \\\ C ( 1 - \alpha ) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C \alpha \ \ \ \ \ \ \ \ \ \ \ \ C \alpha\\\ \end{array}
Kh=KwKb=10−2K_h = \frac{ K_ w }{ K_b } = 10^{ - 2 }
10⇒α2+α−1=0\Rightarrow \alpha^2 + \alpha - 1 = 0
⇒α=−1+1+3020=0.27\Rightarrow \alpha = \frac{ - 1 + \sqrt{ 1 + 3 0 }}{ 2 0 } = 0.27
⇒[H+]=Cα=0.1×0.27=0.027M\Rightarrow [H^+ ] = C \alpha = 0.1 \times 0.27 = 0.027 M