1. for every 5.00 ml of milk of magnesia there are 400. mg of magnesium hydroxide. how many ml of milk of magnesia do we need to neutralize 40.0 ml of 0.500 m hcl? express your answer in ml.

2 uppose 50.0 ml of 0.350 m lithium hydroxide is mixed with 30.0 ml of 0.250 m perchloric acid. what is the ph of the resulting solution, assume the reaction goes to completion

1. 6.50mL

2. pH = 13.097

Explanation:

1. The neutralization of HCl with Al(OH)₃ is:

3HCl + Al(OH)₃ → Al(Cl)₃ + 3H₂O

40.0mL ≡ 0.0400L of 0.500M HCl are:

0.0400L × (0.500mol / L) = 0.0200 moles of HCl

Based on the neutralization, 3 moles of HCl react with 1 mole of Al(OH)₃, thus, the moles of Al(OH)₃ that react with 0.0200 mol of HCl are:

0.0200mol HCl × (1mol Al(OH)₃ / 3mol HCl) = 0.00667 moles of Al(OH)₃. In grams:

0.00667 moles Al(OH)₃ × (78g / 1mol) = 0.520g of Al(OH)₃ ≡ 520mg

As 5.00mL of milk magnesia contain 400mg of Al(OH)₃, the mL of milk magnesia required for a complete reaction are:

520 mg Al(OH)₃ × (5.00mL / 400mg) = 6.50mL

2. The reaction of lithium hydroxide (LiOH) with perchloric acid (HClO₄) is:

HClO₄ + LiOH → LiClO₄ + H₂O

The reaction is 1:1.

Moles of LiOH and HClO₄ are:

LiOH: 0.0500L × (0.350mol / L) = 0.0175 moles of LiOH

HClO₄: 0.0300L × (0.250mol / L) = 0.0075 moles of HClO₄

Assuming the reaction goes to completion, moles of LiOH that remains are:

0.0175 mol - 0.0075 mol = 0.0100 moles of LiOH. The total volume is 80.0mL, 0.0800L. Thus, molarity of LiOH is:

0.0100 mol / 0.0800L = 0.125 M of LiOH

It is possible to obtain the pOH of the solution thus:

pOH = -log (OH) = -log 0.125M = 0.903

As pH = 14- pOH,

pH = 14 - 0.903 = 13.097

I hope it helps!

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