55.00 ml of 1.482 m h2so4 solution and 10.00 ml of 0.269 m naoh solution are combined, and then diluted with water until the total volume is 250.00 ml. what is the molarity of the final h2so4 solution?

Volume of H₂SO₄ solution = 55.00 mL = 55/1000 = 0.055 L

Molarity of H₂SO₄ solution = 1.482 M

Molarity = moles / solution in Liter

Moles of H₂SO₄ = Molarity  x solution in Liter

                      =  1.482 M x 0.055 L

                      = 0.08151

Volume of NaOH solution = 10.00 mL = 10/1000 = 0.01 L

Molarity of NaOH solution = 0.269 M

Moles of NaOH =  0.269 M x 0.01 L = 2.96 x 10⁻³

Now writing a balanced equation for the chemical reaction between H₂SO₄ and NaOH:

H₂SO₄+ 2NaOH --> Na₂SO₄ + 2H₂O

So 2 moles of NaOH reacts with 1 mole of H₂SO₄. NaOH being least in the number of moles will be the limiting reactant.

So 2.96 x 10⁻³ moles of NaOH will neutralize 2.96 x 10⁻³ moles of H₂SO₄.

Unreacted moles of H₂SO₄ = 0.08151 - 2.96 x 10⁻³ = 0.07855

Total volume of the solution = 250 ml = 0.25 L

Thus, the molarity of the final H₂SO₄ solution = moles / Liter of solution

= 0.07855 moles/ 0.25 L = 0.3142 M

when [oh-] is 10-5  m, and [h3o+] is 10-9  m, the poh is 5 and ph is 9. again in both cases the sum of ph and poh is 14. however, the hydrogen ion concentration is not always going to be equal to exactly 1 x 10 raised to a negative number. for example, we skipped over the value of 2.0 x 10-7.

1. they vary with a higher number because boiling is way hotter than melting