9.7 g Bi(NO3)·5H2O and 0.034 g AgNO3 were dissolved in concentrated nitric acid under vigorous stirring at 30°C to form a clear solution. The sample obtained above was labeled as Ag-1.
To determine the molar ratio of Ag/Bi in the solution, you need to calculate the moles of silver nitrate (AgNO3) and bismuth nitrate pentahydrate (Bi(NO3)·5H2O) in the solution.
First, calculate the moles of AgNO3:
0.034 g AgNO3 / 169.87 g/mol = 0.0002 mol AgNO3
Next, calculate the moles of Bi(NO3)·5H2O:
9.7 g Bi(NO3)·5H2O / (357.01 g/mol) = 0.0271 mol Bi(NO3)·5H2O
Finally, divide the moles of AgNO3 by the moles of Bi(NO3)·5H2O to get the molar ratio of Ag/Bi:
0.0002 mol AgNO3 / 0.0271 mol Bi(NO3)·5H2O = 0.00738
So, the molar ratio of Ag/Bi is 0.00738, or 0.738%. To get a 1% molar ratio of Ag/Bi, you would need to add additional AgNO3 to the solution until the ratio reaches 1%.
About silver nitrate aq. sol. acidified with nitric acid ― at this forum: https://www.researchgate.net/post/Why_does_silver_nitrate_look_turbid_after_adding_nitric_acid
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