What is the concentration of NO3- ions when equal volumes of 0.1M AgNO3 and 0.1M NaCl are mixed together?
hi even i hav this doubt
0.1/2=0.05
not clearified
Why finalvol.=2. Initial vol in reaction
To find the concentration of NO3- ions when equal volumes of 0.1M AgNO3 and 0.1M NaCl are mixed together, we need to consider the reaction that occurs.
When AgNO3 and NaCl are mixed, they react to form AgCl and NaNO3 in a double displacement reaction: AgNO3 + NaCl -> AgCl + NaNO3.
Since AgCl is insoluble in water, it precipitates out of solution as a solid. NaNO3, on the other hand, is soluble and remains in solution.
Since the balanced equation shows a 1:1 ratio between AgNO3 and NaCl, and assuming equal volumes are mixed, we can infer that the amount of AgNO3 reacting is equal to the amount of NaCl reacting.
Therefore, if we start with equal volumes of 0.1M AgNO3 and 0.1M NaCl, after mixing, the concentration of NO3- ions is still 0.1M.
In other words, the concentration of NO3- ions remains the same when equal volumes of 0.1M AgNO3 and 0.1M NaCl are mixed together.
(NO3^-) = 0.1/2 = 0.05M. If you don't see that, then assume a volume (any value) of AgNO3 and NaCl and mix them together. For example, 100 mL of each, then
millimoles AgNO3 = 100 x 0.1 = 10
Then M NO3^- = 10mmoles/200 mL = 0.05M.