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18 September, 03:15

The volume of a sample of pure HCl gas was 205 mL at 27°C and 141 mmHg. It was completely dissolved in about 70 mL of water and titrated with an NaOH solution; 24.3 mL of the NaOH solution was required to neutralize the HCl. Calculate the molarity of the NaOH solution.

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  1. 18 September, 03:30
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    The answer to your question is: Molarity = 0.078

    Explanation:

    Data

    HCl

    V = 250 ml

    T = 27°C = 300 °K

    P = 141 mmHg = 0.185 atm

    V2 = 70 ml

    NaOH

    V = 24.3 ml

    Molarity NaOH = ?

    Process

    1. - Calculate the number of moles of HCl

    PV = nRT

    n = PV / RT

    R = 0.082 atm l / mol K

    n = (0.185) (0.25) / (0.082) (300)

    n = 0.046 / 24.6

    n = 0.0019 moles

    2. - Calculate molarity of HCl

    Molarity = moles / volume

    Molarity = 0.0019 / 0.070

    Molarity = 0.027

    3. - Write the balanced equation

    HCl + NaOH ⇒ H₂O + NaCl

    Here, we observe that the proportion HCl to NaOH is 1:1.

    Then 0.0019 moles of HCl reacts with 0.0019 moles of NaOH.

    4. - Calculate the molarity of NaOH.

    Molarity = 0.0019 / 0.0243

    Molarity = 0.078
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