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11 August, 04:08

A 0.105 L sample of an unknown HNO 3 solution required 35.7 mL of 0.250 M Ba (OH) 2 for complete neutralization. What is the concentration of the HNO 3 solution?

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  1. 11 August, 04:10
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    Answer: 0.17M

    Explanation:

    The equation for the reaction is:

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

    From the balanced equation, we obtain:

    nA = mole of acid = 2

    nB = mole of the base = 1

    From the question, we obtain:

    Va = Vol. Of acid = 0.105L

    Ma = conc. Of acid = ?

    Vb = Vol of base = 35.7 mL = 0.0357L

    Mb = conc. of base = 0.25M.

    We solve for the conc. of the acid using:

    MaVa / Mb Vb = nA / nB

    (Ma x 0.105) / (0.25x0.0357) = 2

    Cross multiply to express in linear form. We have:

    Ma x 0.105 = 0.25 x 0.0357 x 2

    Divide both side by 0.105. We have

    Ma = (0.25 x 0.0357 x 2) / 0.105

    Ma = 0.17M

    Therefore the concentration of the acid (HNO3) is 0.17M
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