First write the balanced reaction
2HCl + Mg(OH)2-----------------> MgCl2 + 2HOH
Moles of HCl
Conversion 200ml= 0.2L
=0.2L * 0.150mole/L
= 0.03 moles
Moles of Mg(OH)2
Conversion 125 ml= 0.125L
=0.125L * 0.175mole/L
= 0.021875 moles
Moles of MgCl2 base of HCl
(0.03mol HCl/1)* (1mol MgCl2 / 2 mole HCl)
=0.015 MgCl2
Moles of MgCl2 base of Mg(OH)2
(0.021875 Mg(OH)2 / 1)* (1mol MgCl2 / 1 mole Hg(OH)2)
=0.021875 MgCl2
Since the first is smaller (0.015 mole) than that of the the second (0.0281875) , Therefore, I can conclude that the solution is aciditic.
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