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4 September, 12:02

A 0.080L solution of Ca (OH) 2 is neutralized by 0.0293L of a 3.58 M H2CrO4 solution. What is the concentration of the Ca (OH) 2 solution?

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  1. 4 September, 14:36
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    1.31 M

    Explanation:

    Step 1:

    Data obtained from the question. This include the following:

    Volume of Base (Vb) = 0.080L

    Molarity of base (Mb) = ... ?

    Volume of acid (Va) = 0.0293L

    Molarity of acid (Ma) = 3.58 M

    Step 2:

    The balanced equation for the reaction. This is given below:

    H2CrO4 + Ca (OH) 2 → CaCrO4 + 2H2O

    From the balanced equation above,

    The mole ratio of the acid (nA) = 1

    The mole ratio of the base (nB) = 1

    Step 3:

    Determination of the concentration of base.

    The concentration of the base can be obtained as follow:

    MaVa/MbVb = nA/nB

    3.58 x 0.0293 / Mb x 0.080 = 1

    Cross multiply

    Mb x 0.080 = 3. 58 x 0.0293

    Divide both side by 0.080

    Mb = (3.58 x 0.0293) / 0.08

    Mb = 1.31 M

    Therefore, the concentration of the base, Ca (OH) 2 is 1.31 M
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