20 mL,0.1 M HA [(K_b)_{A^-} = 10^{-10}] is | vedclass

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(C) Given: $HA$ is a weak acid. $(K_b)_{A^-} = 10^{-10}$,so $K_a = K_w / K_b = 10^{-14} / 10^{-10} = 10^{-4}$. Thus,$pK_a = 4$.$1$. Initial $pH$: $[H^

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Phenolphthalein is suitable indicator for this titration but methyl orange is not suitable

Vedclass · Answer

(C) Given: $HA$ is a weak acid. $(K_b)_{A^-} = 10^{-10}$,so $K_a = K_w / K_b = 10^{-14} / 10^{-10} = 10^{-4}$. Thus,$pK_a = 4$.$1$. Initial $pH$: $[H^+] = \sqrt{K_a 	imes c} = \sqrt{10^{-4} 	imes 0.1} = \sqrt{10^{-5}} = 10^{-2.5}$. So,$pH = 2.5$.$2$. Half equivalence point: $pH = pK_a = 4$.$3$. Equivalence point: Total volume = $20 + 10 = 30 \ mL$. $[A^-] = (20 	imes 0.1) / 30 = 2/30 = 1/15 \ M$.$pH = 7 + 0.5(pK_a + \log c) = 7 + 0.5(4 + \log(1/15)) = 7 + 0.5(4 - 1.18) = 7 + 1.41 = 8.41$.$4$. Indicator: Since the $pH$ at equivalence point is $8.41$ (basic range),Phenolphthalein ($pH$ range $8.2-10$) is suitable,while Methyl orange ($pH$ range $3.1-4.4$) is not.

List-$I$	List-$II$
$A$. $Aq.$ solution of $AlCl_3$	$I$. Basic
$B$. $Aq.$ solution of $CH_3COONa$	$II$. Acidic
$C$. $Aq.$ solution of $KCl$	$III$. Highly conductive
$D$. $Al_2O_3$	$IV$. Strongly basic
	$V$. Amphoteric

$20 \ mL$,$0.1 \ M \ HA$ $[(K_b)_{A^-} = 10^{-10}]$ is titrated against $0.2 \ M \ NaOH$. Select the correct statement [Given : $\log 2 = 0.3, \log 3 = 0.48$ ]

Similar Questions

At $60\,^{\circ}C$,the concentration of $[H_3O^{+}]$ in pure water is $10^{-6.7} \, \text{mol/L}$. What is the value of $K_W$ at $60\,^{\circ}C$?

At $25 \, ^\circ C$,the dissociation constant of pure water is equal to .......

Match the following:
List-$I$ List-$II$
$A$. $Aq.$ solution of $AlCl_3$ $I$. Basic
$B$. $Aq.$ solution of $CH_3COONa$ $II$. Acidic
$C$. $Aq.$ solution of $KCl$ $III$. Highly conductive
$D$. $Al_2O_3$ $IV$. Strongly basic
$V$. Amphoteric

The correct match is:

In which of the following acid-base titrations is the $pH$ at the equivalence point greater than $8$?

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