A 1.64 g sample containing a weak acid HX ( MW = 82.0 ) is dissolved in 60 ml water and titrated with 0.250 M NaOH . When half of the acid was neutralised the pH is 5.0 . At the equivalence point the

Question

A 1.64 g sample containing a weak acid HX ( MW = 82.0 ) is dissolved in 60 ml water and titrated with 0.250 M NaOH . When half of the acid was neutralised the pH is 5.0 . At the equivalence point the pH is 9.0. Calculated percentage of HX in sample.

TestHub · Accepted Answer

When half the acid is neutralized, pH = pKa = 5.0. At the equivalence point, moles of $N a O H$ = moles of $H X$ . Moles of $N a O H$ = $V \times M$ . We find V from the titration to pH 5.0. Then, we calculate the moles of $H X$ , convert to grams, and find the percentage:

$p H = p K a = 5.0$

$[H^{+}] = 1 0^{- 5}$

At half neutralization, $[H X] = [X^{-}]$ .

$M o l es o f N a O H = M o l es o f H X n e u t r a l i ze d$

$V o l u m e o f N a O H a t ha l f n e u t r a l i z a t i o n = \frac{M o l es}{M o l a r i t y}$

$M o l es o f H X = 0.250 M \times V o l u m e o f N a O H (L)$

At equivalence, $m o l es H X = m o l es N a O H$ .

$M o l es H X = \frac{1.64 \times p erce n t a g e}{82}$

$P erce n t a g e = 50%$

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