A 100.0 mL portion of 0.250 M calium nitrate solution is mixed with 400.0 mL of .100M nitric acid solution. What is the final concentration of the nitrate ion?
Ca(NO3)2: nitrate ion .5M, so you have .05moles
HNO3: nitrate ion, .1M, so you have .04moles
Nitrate ion concentration=(.09)/.5L=.18M
where did you get the .5M from?
Ca(NO3)2 is .25M
To find the final concentration of the nitrate ion in the solution, we need to calculate the number of moles of nitrate ions present in each solution and then combine them.
First, let's calculate the number of moles of nitrate ions in the 100.0 mL portion of 0.250 M calcium nitrate solution:
Molarity (M) = moles (mol) / volume (L)
Rearranging the equation to solve for moles:
moles (mol) = Molarity (M) * volume (L)
Moles of nitrate ions in the calcium nitrate solution = 0.250 M * 0.100 L = 0.025 mol
Next, let's calculate the number of moles of nitrate ions in the 400.0 mL portion of 0.100 M nitric acid solution:
Moles of nitrate ions in the nitric acid solution = 0.100 M * 0.400 L = 0.040 mol
Now, we need to combine the moles of nitrate ions from both solutions:
Total moles of nitrate ions = moles from calcium nitrate solution + moles from nitric acid solution = 0.025 mol + 0.040 mol = 0.065 mol
To find the final concentration of the nitrate ion, divide the total moles by the total volume of the combined solutions:
Total volume of the combined solutions = 100.0 mL + 400.0 mL = 500.0 mL = 0.500 L
Final concentration of the nitrate ion = Total moles / Total volume = 0.065 mol / 0.500 L = 0.130 M
Therefore, the final concentration of the nitrate ion in the solution is 0.130 M.