Acids & bases/concentrationA mixture is made by mixing 425 mL of 0.94 M H2SO4 with 750. mL of 0.83 M NaOH. (a) What is the limiting reactant? (b) How much excess reactant will be left over? (c) Is this mixture acidic or basic? Why?(d) What is the concentration of the [H+] or [OH-] ions that remain in this solution?(e) Calculate the pH of this solution.

a. NaOH is limiting reactant

b. 0.0882 moles H₂SO₄

c. Acidic

d. 0.15M [H⁺]

e. pH = 0.82

Explanation:

The reaction of NaOH with H₂SO₄ is:

2NaOH + H₂SO₄ → 2H₂O + Na₂SO₄

a. To calculate limiting reactant:

Moles NaOH:

750mL = 0.750L ₓ (0.83mol / L) = 0.6225moles NaOH

Moles H₂SO₄:

425mL = 0.425L ₓ (0.94mol / L) = 0.3995moles H₂SO₄

As 1 mole of H₂SO₄ reacts with 2 moles of NaOH, moles of NaOH that must react with 0.3995moles H₂SO₄ are:

0.3995moles H₂SO₄ ₓ (2moles NaOH / 1 mole H₂SO₄) = 0.799 moles of NaOH.

As you have just 0.6225 moles of NaOH

b. 0.6225 moles of NaOH will react with:

0.6225 moles NaOH ₓ (1 mole H₂SO₄ / 2 mole NaOH) = 0.3113 moles H₂SO₄

That means will remain:

0.3995moles H₂SO₄ - 0.3113 moles H₂SO₄ =

c. As you have an excees of sulfuric ACID, the mixture will be acidic

d. From 1 mole of H₂SO₄ you will have in solution 2 moles of H⁺, thus, from 0.0880 moles of H₂SO₄you will obtain 0.0882*2 = 0.1764 moles of H⁺

In 425 + 750mL = 1175mL = 1.175L

[H⁺] = 0.1764 moles of H⁺ /  1.175L =

e. As pH is -log  [H⁺]

answer: mercury

explanation:

14,16 18 reacts with group