How many mL of 0.406 M NaOH solution are required to completely titrate 32.84 mL of 0.119 M HNO3 solution?

Select one:

a. 9.63

b. 112

c. 680

d. 0.00159

2 Answers

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  • 6 months ago

    Titration ....

    The calculations for a titration are just like the calculations for any other situation involving solutions. Use the unit-factor method (aka dimensional analysis).

    HNO3(aq) + NaOH(aq) --> NaNO3(aq) + HOH(l)

    32.84 mL ......??? mL

    0.119M ......... 0.406M

    0.03284L x (0.119 mol HNO3 / 1L) x (1 mol NaOH / 1 mol HNO3) x (1L / 0.406 mol NaOH) = 0.00963L ... or .... 9.63 mL of NaOH

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  • God
    Lv 7
    6 months ago

    v x M = v x M

    v x 0.406 = 32.84 x 0.119

    v = (32.84 x 0.119)/0.406 = 9.63mL

    Therefore, (a) is correct

    • pisgahchemist
      Lv 7
      6 months agoReport

      NO, NO, NO ..... Just because you've forced the "dilution equation" into a situation to calculate a titration, doesn't make it a good idea. It may work some of the time, but not all of the time.

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