Nitric acid + mg(no3)35.0 ml of 0.255 m nitric acid is added to 45.0 ml of 0.328 m mg(no3)2. what is the concentration of nitrate ion in the final solution?
a. 0.481 m
b. 0.296 m
c. 0.854 m
d. 1.10 m
e. 0.0295 m

The correct answer is option a.

The concentration of nitrate ion in the final solution is 0.48056 M.

Explanation:

Moles of nitrate ions in 35.0 ml of 0.255 M nitric acid:

Moles of nitric acid = 0.008925 mol

1 mol of nitric acid gives 1 mol of nitrate ions,

Then 0.008925 moles of nitric acid will give 0.008925 moles of nitrate ions.

Moles of nitrate ions in 35.00 mL of solution = 0.008925 mole ..(1)

Moles of nitrate ions in 45.0 ml of 0.328 M magnesium nitrate:

Moles of magnesium nitrate = 0.01476 mol

1 mol of magnesium nitrate gives 2 mol of nitrate ions,

Then 0.01476 moles of magnesium nitrate will give :

of nitrate ions

Moles of nitrate ions in 45.00 mL of solution = 0.02952 mol...(2)

Total number of moles of nitarte ions =

0.008925 mol+0.02952 mol = 0.038445 mol

Total volume after mixing = 35.00 mL+ 45.00 ml =

80.00 mL = 0.080 L

The concentration of nitrate ion in the final solution:

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