Question

3. A primary standard is a reagent that is extremely pure, stable, has no waters of hydration, and typically has a high molecular weight. One such stable standard, even though its formula weight is not high, is sulfamic acid, NH SOOH (formula weight 97.09 g/mol). How many grams of this salt will be required to titrate 50.0 ml of 0.15 M NaOH solution? NH3SOOH (aq) + NaOH(aq) - NH2SOONa (aq) + H2O (1) *1314-YEAR 2019-2020 11-12

Answer 1

volume NaOH = 50 ml =0.050 L

molarity of NaOH =0.15 M

so moles of NaOH present = moarity*volume = 0.15 M *0,050 L = 0.0075 moles

balanced equation says 1 mole sulfamic acid required to neutralize 1 mole NaOH

thus moles of acid needed=0.0075 moles

mole mass=97.09 g/mol

Hence grams of acid present = moles of acid required *molar mass = 0.0075 moles*97.09 g/mol

=0.728175 g

after rounding

mass needed = 0.728 g

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Thanks