9-125 g of HCl is present in 0-5 L of solution at 25-C- The pH of the solution is -

Volume =0.5 L Mass of HCl = 9.125 g Molecular mass of HCl=36.5 g mol^ 1 Concentration=MolesVolume=Given MassMolecular mass× Volume Concentration=9.12536.5×0 ...

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Byju's Answer

Standard XIII

Chemistry

pH the Power of H

9.125 g of HC...

Question

9.125 g of $H C l$ is present in 0.5 L of solution at $25^{\circ} C$ . The pH of the solution is :

1.44

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0.88

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0.6

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0.3

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Solution

The correct option is D 0.3
Volume =0.5 L
Mass of $H C l = 9.125 g$
Molecular mass of $H C l = 36.5 g m o l^{- 1}$
$C o n c e n t r a t i o n = \frac{M o l e s}{V o l u m e} = \frac{G i v e n M a s s}{M o l e c u l a r m a s s \times V o l u m e} C o n c e n t r a t i o n = \frac{9.125}{36.5 \times 0.5} m o l L^{- 1} = \frac{1}{2} M$
The dissociation of HCl in aqueous medium is given by : $H C l + H_{2} O ⟶ H_{3} O^{+} + C l^{-} I n i t i a l : C 00 E q u i l i b r i u m : 0 C C$ .

$[H C l]_{i n i t i a l l y} = [H^{+}]_{e q b .} = C [H^{+}] = 0.5 M$
$p H = - l o g [H^{+}] = - l o g (0.5) = l o g 2 p H = 0.3$

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pH the Power of H

Standard XIII Chemistry

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9.125 g of HCl is present in 0.5 L of solution at 25∘C. The pH of the solution is :

9.125 g of $H C l$ is present in 0.5 L of solution at $25^{\circ} C$ . The pH of the solution is :