1.

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)

Answer»

-0.176
`+0.477`
-0.2345
0.3

Solution :`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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