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?
BOB PURSLEY PLEASE EXPLAIN MORE THOROUGHLY HOW YOU GOT THE NITRATE ION TO BE .5M
Well, you are right. The Ca(NO3)2 IS 0.25 M just as you said; however, the problem asks for NO3^- and since there are two moles NO3^- per 1 mole Ca(NO3)2, that makes the NO3^- just twice or 0.250 x 2 = 0.5 M and you took 100 mL of that so M x L = moles = 0.05 moles just like Bob P wrote.
thank doctor bob pursley that's right i know who you are.
No. I can guarantee you that Bob P and DrBob222 are NOT the same person. In fact, he and I live about 400 miles from each other. I just picked up your confusion over the nitrate ion because he wasn't on-line at the moment.
o oops hahah thankss for answering my question!
1)How many milliliters of 1.5 M NaOH will react completely with 448 mL of 0.600 M HCl?
NaOH (aq) + HCl (aq) �¨ NaCl (aq) + H2O (l)
2)What is the percent yield of the following reaction if 15 g H2O are obtained from the combustion of 18 g C2H6?
2 C2H6 (g) + 7 O2 (g) �¨ 4 CO2 (g) + 6 H2O (g)
3)What is the molarity of a solution in which 6.4 g HCl is dissolved in enough water to make 2.4 L solution?
4)How many atoms of H are there in 4.3 g CH4?
I REALLY NEED HELP SOLVING THESE CHEM PROBLEMS
To find the final concentration of the nitrate ion, we need to use the concept of mole-to-mole ratio and the volume of the solutions.
Let's start by calculating the number of moles of nitrate ions present in the 100.0 mL of 0.250 M calcium nitrate solution:
moles of nitrate ions in calcium nitrate solution = concentration × volume
= 0.250 mol/L × 0.100 L
= 0.025 mol
Now, let's calculate the number of moles of nitrate ions present in the 400.0 mL of 0.100 M nitric acid solution:
moles of nitrate ions in nitric acid solution = concentration × volume
= 0.100 mol/L × 0.400 L
= 0.040 mol
Next, we need to consider the reaction between calcium nitrate (Ca(NO3)2) and nitric acid (HNO3):
Ca(NO3)2 + 2HNO3 -> 2HNO3 + Ca(NO3)2
From the balanced equation, we can see that for every one mole of calcium nitrate, two moles of nitrate ions are produced.
Therefore, the total moles of nitrate ions after the reaction is carried out will be:
total moles of nitrate ions = moles of nitrate ions in calcium nitrate solution + moles of nitrate ions in nitric acid solution
= 0.025 mol + 0.040 mol
= 0.065 mol
Now, we need to calculate the total volume of the final solution. The volume of the calcium nitrate solution is 100.0 mL, and the volume of the nitric acid solution is 400 mL. Therefore, the total volume is:
total volume = volume of calcium nitrate solution + volume of nitric acid solution
= 100.0 mL + 400.0 mL
= 500.0 mL
Now, we can calculate the final concentration of the nitrate ion using the formula:
final concentration = total moles of nitrate ions / total volume
final concentration = 0.065 mol / 0.500 L
final concentration = 0.130 M
So, the final concentration of the nitrate ion in the solution is 0.130 M.