SATHEE: Chemical and Ionic Equilibrium 2 Question 53

Chemical and Ionic Equilibrium 2 Question 53

53. $0.1 M$ of $H A$ is titrated with $0.1 M NaOH$ , calculate the $pH$ at end point. Given, $K_{a} (H A) = 5 \times 10^{- 6}$ and $α « 1$ .

(2004)

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Solution:

At the end-point, $[A^{-}] = 0.05$

$\begin{aligned} K_{b} (A^{-}) & = K_{w} / K_{a} = 2 \times 10^{- 9} [{OH}^{-}] & = \sqrt{K_{b} C} = \sqrt{2 \times 10^{- 9} \times 0.05} = 10^{- 5} pOH & = 5 and pH = 9 \end{aligned}$

Chemical and Ionic Equilibrium 2 Question 53

53. $0.1 M$ of $H A$ is titrated with $0.1 M NaOH$ , calculate the $pH$ at end point. Given, $K_{a} (H A) = 5 \times 10^{- 6}$ and $α « 1$ .

Solution:

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Chemical and Ionic Equilibrium 2 Question 53

53. 0.1M of HA is titrated with 0.1MNaOH, calculate the pH at end point. Given, Ka(HA)=5×10−6 and «α«1.

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

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53. $0.1 M$ of $H A$ is titrated with $0.1 M NaOH$ , calculate the $pH$ at end point. Given, $K_{a} (H A) = 5 \times 10^{- 6}$ and $α « 1$ .