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In Lab H2SO4 solution is prepared by dissolving 4.9 grams of H2SO4 in 100mL of water a. How many h2SO4 molecules are in this solution b. What is the molarity of the soltuion c. What is the pH of the solution d. What is the pH of the solution if 3.0 grams of NaOH is added to the solution

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Answer:

Part (a) The number of molecules of H₂SO₄ in the solution is 3.011 X 10²² molecules

Part (b) the molarity of the solution is 0.5 M

Part (c) the pH of the solution is 0

Part(d) the pH of the solution is 13.87

Explanation:

Part (a) How many H₂SO₄ molecules are in this solution

number of moles = Reacting mass/molar mass

Molar mass of H₂SO₄ = (2x1 + 32 + 4X16) = 98 g/mol

      number of moles = 4.9/98 = 0.05 mol.

I mole =  6.022 x 10²³ molecules

0.05 mole = (0.05 X 6.022 x 10²³) molecules

= 3.011 X 10²² molecules

Part (b) What is the molarity of the solution

molarity = moles of solute/liters of solution

molarity = 0.05/(100 x 10⁻³ L)

molarity = 0.5 M

Part (c) What is the pH of the solution

H₂SO₄ ⇄ 2H⁺ + SO₄²⁻

pH = -Log[2H⁺]

pH = -Log[2*0.5]

pH = -Log[1]

pH = 0

Part(d) What is the pH of the solution if 3.0 grams of NaOH is added to the solution

molar mass of NaOH = ( 23 + 16 +1) = 40g/mol

Number of moles = 3/40 = 0.075 moles

molarity = 0.075/(100 x 10⁻³ L)

molarity = 0.75 M

pOH = - Log[OH⁻]

pOH = - Log[0.75]

pOH = 0.13

Also, pH + pOH = 14

pH = 14 - pOH

pH = 14 - 0.13

pH = 13.87

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