2.5 mL of (2/5) M weak monoacidic base (Kb=1×10−12 at 25∘) is titrated (2/15) M HCl in water at 25∘C. The concentration of H+ at equivalence point is (Kw=1×10−14 at 25∘C)

mmol of base =2.5×52=1mmol of acid required to reach the end point =1Volume of acid required to reach the end point =215 mLTotal volume at the end point =215+2.5=10 mLMolarity of salt at the end point =101=0.10C(1−a)B++H2O⇌CaBOH+CaH+Kh=KbKw=10−2Kh=10−2=1−aCa=1−a0.1a⇒10a2+a−1=0⇒a=20−1+1+40=0.27⇒[H+]=Ca=0.1×0.27=0.027M

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The acid base titration using indicators

2.5 {~mL} of (2 / 5) {~M~} weak monoacidic base (K_{b}=1 10^{

Question

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Solving time: 5 mins

$2.5 mL$ of $(2/5) M$ weak monoacidic base $(K_{b} = 1 \times 1 0^{- 12}$ at $2 5^{\circ})$ is titrated $(2/15) M HCl$ in water at $2 5^{\circ} C$ . The concentration of $H^{+}$ at equivalence point is $(K_{w} = 1 \times 1 0^{- 14}$ at $2 5^{\circ} C)$

3.7 \times 1 0^{- 14} M

3.2 \times 1 0^{- 7} M

3.2 \times 1 0^{- 2} M

2.7 \times 1 0^{- 2} M

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Updated on: Oct 14, 2023

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Theodore Discussed

2.5 mL

(2/5) M

weak monoacidic base

(K_{b} = 1 \times 1 0^{- 12}

2 5^{\circ})

is titrated

(2/15) M HCl

in water at

2 5^{\circ} C

. The concentration of

H^{+}

at equivalence point is

(K_{w} = 1 \times 1 0^{- 14}

2 5^{\circ} C)

12 mins ago

Discuss this question LIVE

12 mins ago

Text solutionVerified

mmol of base

= 2.5 \times \frac{2}{5} = 1

mmol of acid required to reach the end point

= 1

Volume of acid required to reach the end point

= \frac{15}{2} mL

Total volume at the end point

= \frac{15}{2} + 2.5 = 10 mL

Molarity of salt at the end point

= \frac{1}{10} = 0.10

C (1 - a) B^{+} + H_{2} O ⇌ Ca BOH + Ca H^{+}

K_{h} = \frac{K _{w}}{K _{b}} = 1 0^{- 2}

K_{h} = 1 0^{- 2} = \frac{Ca}{1 - a} = \frac{0.1 a}{1 - a}

\Rightarrow 10 a^{2} + a - 1 = 0

\Rightarrow a = \frac{- 1 + 1 + 40}{20} = 0.27

\Rightarrow [H^{+}] = Ca = 0.1 \times 0.27 = 0.027 M

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Question Text	$2.5 mL$ of $(2/5) M$ weak monoacidic base $(K_{b} = 1 \times 1 0^{- 12}$ at $2 5^{\circ})$ is titrated $(2/15) M HCl$ in water at $2 5^{\circ} C$ . The concentration of $H^{+}$ at equivalence point is $(K_{w} = 1 \times 1 0^{- 14}$ at $2 5^{\circ} C)$
Updated On	Oct 14, 2023
Topic	Equilibrium
Subject	Chemistry
Class	Class 11
Answer Type	Text solution:1 Video solution: 7
Upvotes	751
Avg. Video Duration	6 min

$2.5 mL$ of $(2/5) M$ weak monoacidic base $(K_{b} = 1 \times 1 0^{- 12}$ at $2 5^{\circ})$ is titrated $(2/15) M HCl$ in water at $2 5^{\circ} C$ . The concentration of $H^{+}$ at equivalence point is $(K_{w} = 1 \times 1 0^{- 14}$ at $2 5^{\circ} C)$

Text solutionVerified

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2.5 mL of (2/5) M weak monoacidic base (Kb​=1×10−12 at 25∘) is titrated (2/15) M HCl in water at 25∘C. The concentration of H+ at equivalence point is (Kw​=1×10−14 at 25∘C)

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$2.5 mL$ of $(2/5) M$ weak monoacidic base $(K_{b} = 1 \times 1 0^{- 12}$ at $2 5^{\circ})$ is titrated $(2/15) M HCl$ in water at $2 5^{\circ} C$ . The concentration of $H^{+}$ at equivalence point is $(K_{w} = 1 \times 1 0^{- 14}$ at $2 5^{\circ} C)$