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Question

1.458 g of Mg reacts with 80.0 mL of HCl solution whose pH is 0.477. The change in pH after all Mg has reacted is:

(Assume constant volume. Mg=24.3 g mol1 and log3=0.477)

A
0.176
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B
+0.477
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C
0.2385
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D
+0.3
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Solution

The correct option is A +0.3
Mg(aq)+2HCl(aq)MgCl2(aq)+H2
Number of moles of Mg is 1.45824.3=0.06 moles =60 M moles
Molarity of HCl=3M
3×80=240 M moles
HCl left after reaction= 240(60×2)=120 M moles
Hence molarity of solution= 12080=1.5M
pH=log[H+]=log1.5=0.176
Change in pH=0.176(0.477)=0.3

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