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14 May, 16:33

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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  1. 14 May, 20:10
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    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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