(a) How many grams of KNO3 are needed to make 430 mL of a solution that is to contain 2.9 mg/mL of potassium ion?
2.9 mg/mL x 430 mL = ? mg K^+ needed.
? mg K^+ x (molar mass KNO3/atomic mass K) = mg KNO3 needed.
mg KNO3 x (1 g/1000 mg) = g KNO3 needed.
To determine the number of grams of KNO3 needed to make a solution with a specific concentration of potassium ion, you need to use the molarity (M) and the molecular weight of KNO3. Here's how you can calculate it step by step:
Step 1: Convert the desired concentration of potassium ion from mg/mL to g/L.
The given concentration is 2.9 mg/mL. To convert this to g/L, divide by 1000:
2.9 mg/mL ÷ 1000 = 0.0029 g/mL.
Step 2: Convert the volume of the solution from mL to L.
The given volume is 430 mL. To convert this to L, divide by 1000:
430 mL ÷ 1000 = 0.43 L.
Step 3: Use the molarity formula to calculate the moles of potassium ion.
Molarity (M) is defined as moles (n) of solute divided by the volume (V) of the solution in liters (L).
M = n/V.
Rearranging the formula gives n = M × V.
In this case, n represents the moles of potassium ion, M is the desired concentration in mol/L, and V is the volume of the solution in L:
n = (0.0029 g/mL) × (0.43 L) = 0.001247 g.
Step 4: Calculate the number of grams of KNO3 needed.
The molecular weight of KNO3 is determined by adding the atomic weights of its constituents:
K: 39.10 g/mol
N: 14.01 g/mol
O: 16.00 g/mol (3 oxygen atoms in total)
Adding these values, we get:
39.10 + 14.01 + (16.00 × 3) = 101.10 g/mol.
To find the number of grams of KNO3 needed, divide the moles of potassium ion by the molar mass of KNO3:
n (g) = (0.001247 g) × (101.10 g/mol) = 0.126 g.
Therefore, approximately 0.126 grams of KNO3 are needed to make 430 mL of a solution that contains 2.9 mg/mL of potassium ion.