Suppose of copper (II) nitrate is dissolved in of a aqueous solution of sodium chromate.
Calculate the final molarity of nitrate anion in the solution. You can assume the volume of the solution doesn't change when the copper (II) nitrate is dissolved in it.
Round your answer to significant digits.
You omitted the amount of copper nitrate and sodium chromate.
To find the final molarity of nitrate anion in the solution, we need to calculate the moles of nitrate anion present and divide it by the final volume of the solution.
First, let's calculate the moles of nitrate anion (NO3-) present in the copper (II) nitrate solution.
Given:
Volume of copper (II) nitrate solution = V1 = 125 mL = 0.125 L
Molarity of copper (II) nitrate = M1 = 0.25 M
Using the formula:
moles = molarity * volume
moles of copper (II) nitrate = M1 * V1
= 0.25 M * 0.125 L
= 0.03125 moles
From the chemical formula of copper (II) nitrate (Cu(NO3)2), we can see that there are 2 nitrate ions per 1 copper (II) ion.
Therefore, moles of nitrate anion in the copper (II) nitrate solution = 2 * 0.03125 moles
= 0.0625 moles
Next, let's calculate the final volume of the solution after the addition of sodium chromate solution.
Given:
Volume of sodium chromate solution = V2 = 250 mL = 0.25 L
The final volume of the solution is the sum of the volumes of the copper (II) nitrate and sodium chromate solutions.
Final volume = V1 + V2
= 0.125 L + 0.25 L
= 0.375 L
Now, let's calculate the final molarity of nitrate anion in the solution.
Molarity = moles / volume
Final molarity of nitrate anion = 0.0625 moles / 0.375 L
= 0.1667 M
Rounding to 3 significant digits, the final molarity of nitrate anion in the solution is 0.167 M.
To find the final molarity of nitrate anion (NO3-) in the solution, we need to calculate the moles of nitrate anion and then divide it by the volume of the solution.
First, let's calculate the moles of nitrate anion in copper (II) nitrate. Copper (II) nitrate is composed of one copper ion (Cu2+) and two nitrate ions (NO3-).
Given:
Amount of copper (II) nitrate = 0.40 moles
Moles of nitrate anion in copper (II) nitrate = 2 * 0.40 = 0.80 moles
Now, let's calculate the final volume of the solution. We're assuming the volume doesn't change when the copper (II) nitrate is dissolved, so it remains at 1.00 L.
Now, use the moles of nitrate anion (0.80 moles) and the volume of the solution (1.00 L) to calculate the final molarity.
Molarity (M) = Moles of solute / Volume of solution
Molarity of nitrate anion = 0.80 moles / 1.00 L = 0.80 M
Note that we have rounded the answer to 2 significant digits, since both the moles and volume are given with two significant digits.
Therefore, the final molarity of nitrate anion in the solution is 0.80 M.