J.R. S. answered 02/20/18
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From pH = 1.3, you can calculate the [H+] as follows:
pH = -log [H+] and thus [H+]= 1x10-1.3 = 5.01x10-2 M
Now, convert this to moles of H+ present in 352 mls (0.352 L)
5.01x10-2 moles/L x 0.352 L = 1.76x10-2 moles H+
How much Mg(OH)2 is present?
175 mg Mg(OH)2 x 1 mmole/58.3 mg = 3.00 mmoles
Since 1 mole Mg(OH)2 produces 2 moles OH-, 3.00 mmoles Mg(OH)2 will neutralize 6.00 mmoles H+
1.76x10-2 moles H+ x 1000 mmoles/mole = 17.6 mmoles H+ present
6.00 moles/17.6 mmoles (x100%) = 34.1% neutralized