50.00mL of 0.1500M NaNO3 (aq) was added to a sample containing solid gallium. Ni

Question

50.00mL of 0.1500M 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 27.82mL of 0.100M NaOH. Determine the mass of Ga(s) in the sample.

3NO- 3   +  9H+  + 6e-  -->  3HNO2  +  3H2O   E= 0.940V

                          2 Ga(s)  -->  2Ga3+  +  6e-        E= -0.549V

Balanced Net Eq: 3NO-3  +  9H+  +  2Ga(s)  -->  3HNO2  +  3H2O  +  2Ga2+

HNO2(aq) <--> H+(aq)  + NO- 2(aq)   Ka=7.1 x 10^-4

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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