Question

What volume (in mL) of 0.750 M sodium hydroxide is needed to neutralize (react with) 275 ml of 0.500 M sulfuric acid (H2SO4)?

I NEED WORK

Answer 1

`\\ \rm\Rrightarrow C_1V_1=C_2V_2`

`\\ \rm\Rrightarrow 0.750V_1=275(0.5)`

`\\ \rm\Rrightarrow 3/4V_1=137.5/tex]</p><p>[tex]\\ \rm\Rrightarrow 3V_1=550`

`\\ \rm\Rrightarrow V_1=183.3mL`

Answer 2

To neutralize 275 mL of 0.500 M sulfuric acid, one would need 366.7 mL of 0.750 M sodium hydroxide according to stoichiometry and the balanced chemical equation for the reaction.

Step-by-step explanation:

The question asks for the volume of sodium hydroxide (NaOH) required to neutralize a given volume of sulfuric acid (H₂SO₄). We can solve this using stoichiometry and the balanced chemical equation which, for the reaction between sodium hydroxide and sulfuric acid, is:

H₂SO₄ (aq) + 2NaOH (aq) → Na₂SO₄ (aq) + 2H₂O (1)

The balanced equation shows that two moles of NaOH react with one mole of H₂SO₄. We are starting with 275 mL of 0.500 M H₂SO₄, so let's find out how many moles of H₂SO₄ we have:

1. Moles of H₂SO₄ = Molarity of H₂SO₄ * Volume of H₂SO₄ in liters = 0.500 mol/L * 0.275 L = 0.1375 mol H₂SO₄.
2. Now, remember the mole ratio from the balanced equation is 1:2, meaning you need twice as many moles of NaOH. Thus, we need 0.1375 mol H₂SO₄ * 2 = 0.275 mol NaOH.
3. Finally, we can find the required volume of NaOH by using its molarity: Volume of NaOH = Moles of NaOH / Molarity of NaOH = 0.275 mol / 0.750 M = 0.3667 L, which is 366.7 mL.

Therefore, the volume of 0.750 M sodium hydroxide needed to neutralize 275 mL of 0.500 M sulfuric acid is 366.7 mL.