50.00 mL of 0.1500 M NaNO 3 (aq) was added to a sample containing solid gallium.

Question

50.00 mL of 0.1500 M NaNO3 (aq) was added to a sample containing solid gallium. Nitrate oxidized Ga (s) to Ga3+ (aq). The resulting nitrous acid was titrated to the equivalence point with 27.82 mL of 0.1000 M NaOH (aq). Determine the mass of Ga (s) in the sample.

NO3- (aq) + 3 H+ (aq) + 2 e- --> HNO2 (aq) + H2O                 Eo = 0.940 V

Ga3+ (aq) + 3 e- --> Ga (s)                                                     Eo = -0.549 V

HNO2 (aq) <---> H+ (aq) + NO2- (aq)                                     Eo = 7.1 x 10-4 V

Explanation / Answer

2 Ga + 3 NO3- + 9 H+ => 2 Ga3+ + 3 HNO2 + 3 H2O

HNO2 + NaOH => NaNO2 + H2O


Moles of HNO2 = moles of NaOH = volume x concentration of NaOH

= 27.82/1000 x 0.100 = 0.002782 mol


Moles of Ga = 2/3 x moles of HNO2

= 2/3 x 0.002782 = 0.0018547 mol


Mass of Ga = moles x molar mass of Ga

= 0.0018547 x 69.72

= 0.129 g

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