How many grams of sodium hydroxide are needed to completely neutralize 25.0g of sulfuric acid to form sodium sulfate and water?

Question

8ItItHow many grams of sodium hydroxide are needed to completely neutralize 25.0g of

sulfuric acid to form sodium sulfate and water?8. How many grams of sodium hydroxide are needed to completely neutralize 25.0g of

sulfuric acid to form sodium sulfate and water?8. How many grams of sodium hydroxide are needed to completely neutralize 25.0g of

sulfuric acid to form sodium sulfate and water?

Answer 1

To completely neutralize an acid, there must be one molar equivalent of base for every hydrogen atom produced. In this scenario, the first step is to find the number of moles of sulfuric acid. To do this, we divide the mass given by the molar mass of H₂SO₄ like so:

moles of H₂SO₄ = (g of H₂SO₄)/(MM H₂SO₄)

moles of H₂SO₄ = (25.0)/(98.08)

moles of H₂SO₄ = .255 moles

Then, to find the moles of hydrogen, we must multiply this by 2 since there are 2 moles of hydrogen in every sulfuric acid molecule. As such we are left with .510 moles of hydrogen. Since we know that the moles of hydrogen are equal to the moles of NaOH, we can then surmise that we need to add .510 moles of NaOH to neutralize this amount of sulfuric acid. Thus, our last step in this process is to multiply this number by the molar mass to get the final mass of NaOH needed to neutralize this sulfuric acid:

g of NaOH = (moles of NaOH)(MM of NaOH)

g of NaOH = (.510)/(40.0)

g of NaOH = 20.4

With that answer, we now know we need 20.4 grams of NaOH to neutralize 25.0 g of H₂SO₄.