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15 June, 02:43

Given a balanced chemical equation between H2SO4 (aq) and KOH (aq) H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l) What volume (in mL) of 0.43 M H2SO4 (aq) solution is necessary to completely react with 120 mL of 0.35 M KOH (aq) ? Note: (1) The unit of volume of H2SO4 (aq) is in mL (2) Insert only the numerical value (integer) of your answer (do not include the units or chemical in your answer).

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Answers (2)
  1. 15 June, 03:19
    0
    48.84mL

    Explanation:

    H2SO4 + 2KOH → K2SO4 + 2H2O

    From the question:

    nA = 1

    nB = 2

    From the question given we obtained the following information:

    Ma = 0.43M

    Va = ?

    Mb = 0.35M

    Vb = 120mL

    Using MaVa / MbVb = nA/nB, we can easily find the volume of the acid required. This is illustrated below:

    MaVa / MbVb = nA/nB

    0.43 x Va / 0.35 x 120 = 1/2

    Cross multiply to express in linear form

    2 x 0.43 x Va = 0.35 x 120

    Divide both side by the (2 x 0.43)

    Va = (0.35 x 120) / (2 x 0.43)

    Va = 48.84mL

    Therefore, the volume of H2SO4 required is 48.84mL
  2. 15 June, 04:28
    0
    The volume of the H2SO4 solution is 48.8 mL

    Explanation:

    Step 1: Data given

    Molarity of H2SO4 solution = 0.43 M

    Volume of KOH = 120 mL

    Molarity KOH = 0.35 M

    Step 2: The balanced equation

    H2SO4 (aq) + 2 KOH (aq) → K2SO4 (aq) + 2 H2O (l)

    Step 3: Calculate the volume of H2SO4

    b*Ca*Va = a*Cb*Vb

    ⇒ b = the coefficient of KOH = 2

    ⇒ Ca = the molarity of H2SO4 = 0.43 M

    ⇒ Va = the volume of H2SO4 = ?

    ⇒ a = the coefficient of H2SO4 = 1

    ⇒ Cb = the molarity of KOH = 0.35 M

    ⇒ Vb = the volume of KOH = 0.120 L

    2*0.43 * Va = 1*0.35 * 0.120 L

    0.86 Va = 0.042

    Va = 0.0488 L = 48.8 mL

    The volume of the H2SO4 solution is 48.8 mL
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