Answer :

The volume of aluminum hydroxide required for neutralization is 24.30 mL.

Acid and base:

The compounds which gives H+ ions in the aqueous solution are called acids while those which gives OH- ions are called base.

Neutralization reaction:

It is a chemical reaction that involves a reaction between the equal molar amount of acid and base and form a solution that has pH = 7.

Calculations:

The reaction between aluminum hydroxide and nitric acid is expressed as:

Al(OH)2 + 2HNO3 ----> Al(NO3)2 + 2H2O

The number of moles of nitric acid is calculated as:

Moles of HNO3 = 0.175 M x (30.00/1000) L = 0.00525 mol

The number of moles of Al(OH)2 required for the neutralization is:

Moles of Al(OH)2 = 0.00525 mol/2 = 0.002625 mol

The volume of Al(OH)2 is calculated as:

Volume = 0.002625 mol/0.108 M = 0.0243 L = 24.30 mL

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