if 20.15 ml of 0.150 m ammonium hydroxide is required to neutralize 13.5 ml of sulfurous acid, what is the molarity of the acid?
h2so3(aq)+2nh4oh(aq)→(nh4)2so3(aq)+ 2h2o(l)

The molarity of the acid is 0.112 M

Explanation:

In the reaction of neutralization, 2 moles of NH₄OH react with 1 mol H₂SO₃. Then, in the equivalence point, the number of moles of NH₄OH will be twice the number of moles of H₂SO₃.

In the equivalence point:

moles H₂SO₃ = 1/2 * moles NH₄OH

Then, let´s calculate the number of moles of NH₄OH consumed in the reaction:

number of moles = molar concentration * volume (l)

number of moles NH₄OH = 0.150 mol/l * 0.02015 l = 3.02 x 10⁻³ mol  

Then, the number of moles of H₂SO₃ will be:

moles H₂SO₃ = 1/2 * moles NH₄OH

moles H₂SO₃ = 1/2 * 3.02 x 10⁻³ mol  

moles H₂SO₃ = 1.51 x 10⁻³ moles

Since this number of moles was initially in 13. 5 ml, the concentration of H₂SO₃  will be:

molar concentration = number of moles / volume (l)

molar concentration =  1.51 x 10⁻³ moles / 0.0135 l = 0.112 M  

Have a nice day!

first one is the correct answer

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