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

3NaNO3 + 6 H =3 HNO2 (aq) +3 H2O + 3Na

2Ga3+ (aq) + 6 e- --> 2Ga (s)

27.8*(0.1/1000) H+ IS THERE

In this situation 3HNO2 react with 2Ga

so 27.8*(0.1/1000) H+ are unreacted from 50*(0.15/1000) so 0.005646molar H+ reacted

with 0.005646*2/3 Ga3+ to form 0.005646*2/3 Ga = 0.003764molar Ga

mass will be = 0.003764* M = 0.003764*69.7 = 0.2623g

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