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A 1.458 g of Mg reacts with 80.0 ml of a...

A 1.458 g of Mg reacts with 80.0 ml of a HCl solution whose pH is -0.477.The change in pH when all Mg has reacted.(Assume constant volume. `Mg=24.3g//"mol".`)(log 3=0.47, log2= 0.301)

A

-0.176

B

`+0.477`

C

-0.2345

D

0.3

Text Solution

Verified by Experts

The correct Answer is:
D
`Mg(aq)+2HCl(aq)toMgCl_2(aq)+H_2`
`(1.458 g)/(24.3 g)`
millimoles of HCl =`3xx80=240` millimoles
millimoles of HCl after reaction =240-60x2=120
New Molarity `=120/80=1.5 M`
`pH=-log[H^+]=-log1.5=-0.176`
Change is pH=-0.176-(-0.477)=0.3
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