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19 November, 10:16

what volume of a 0.155 M calcium hydroxide solution is required to neutralize 28.8 mL of a 0.106 M nitric acid?

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  1. 19 November, 10:29
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    9.85mL

    Explanation:

    First, let us write a balanced equation for the reaction. This is illustrated below:

    Ca (OH) 2 + 2HNO3 - > Ca (NO3) 2 + 2H2O

    From the balanced equation above,

    nA (mole of the acid) = 2

    nB (mole of the base) = 1

    Data obtained from the question include:

    Vb (volume of the base) = ?

    Mb (Molarity of base) = 0.155 M

    Va (volume of the acid) = 28.8 mL

    Ma (Molarity of acid) = 0.106 M

    Using MaVa/MbVb = nA/nB, the volume of calcium hydroxide (i. e the base) can be obtain as follow:

    MaVa/MbVb = nA/nB

    0.106 x 28.8 / 0.155 x Vb = 2/1

    Cross multiply to express in linear form as shown below:

    2 x 0.155 x Vb = 0.106 x 28.8

    Divide both side by 2 x 0.155

    Vb = (0.106 x 28.8) / (2 x 0.155)

    Vb = 9.85mL

    Therefore, the Volume of calcium hydroxide is 9.85mL
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