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One liter of approximetely 0.25N HNO3 has been prepared. Upon titration it was found that 5.0 ml of the acid required 11.0 ml of 0.10N NaOH for neutrality. How much concetrated HNO3 at 15.8N must be added to one liter to make it exactly 0.25N?

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    No. of moles of NaOH = 0.011*0.1 = 0.0011M
    Therefore no. of moles of HNO3 in 5ml.
    = 0.0011M
    Concentration of diluted HNO3
    = 0.0011/0.005=0.22N

    Volume of 15.8N HNO3 required = x litre.

    0.22N*1 litre+15.8x = 0.25(1+x)
    Solve for x to get
    =0.00193 l.
    =1.93 ml.

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