A solution contains, in each 5 mL, 0.5 g of potassium acetate (C2H3KO2, MW = 98), 0.5 g of potassium bicarbonate (KHCO3, Z = 100), and 0.5 g of potassium citrate (C6H5K3O7 . H2O, MW = 324). How many milliequivalents of potassium (K+) are represented in each 5 mL of the solution?
To calculate the milliequivalents of potassium (K+) in each 5 mL of the solution, we need to find the moles of potassium in each compound and then convert them to milliequivalents.
First, let's find the moles of potassium in each compound:
1. Potassium acetate (C2H3KO2):
- The molar mass of potassium acetate is 98 g/mol.
- The mass of potassium in each 5 mL is 0.5 g.
- To find the moles, we divide the mass by the molar mass: 0.5 g / 98 g/mol = 0.0051 mol.
2. Potassium bicarbonate (KHCO3):
- The molar mass of potassium bicarbonate is 100 g/mol.
- The mass of potassium in each 5 mL is 0.5 g.
- To find the moles, we divide the mass by the molar mass: 0.5 g / 100 g/mol = 0.005 mol.
3. Potassium citrate (C6H5K3O7.H2O):
- The molar mass of potassium citrate is 324 g/mol.
- The mass of potassium in each 5 mL is 0.5 g.
- To find the moles, we divide the mass by the molar mass: 0.5 g / 324 g/mol = 0.0015 mol.
Now, let's calculate the milliequivalents of potassium (K+) in each 5 mL:
1. Potassium acetate:
- Each mole of potassium (K+) contributes 1 milliequivalent, so 0.0051 mol of potassium gives 0.0051 milliequivalents.
2. Potassium bicarbonate:
- Each mole of potassium (K+) contributes 1 milliequivalent, so 0.005 mol of potassium gives 0.005 milliequivalents.
3. Potassium citrate:
- Each mole of potassium (K+) contributes 1 milliequivalent, so 0.0015 mol of potassium gives 0.0015 milliequivalents.
Finally, let's add up the milliequivalents from each compound:
0.0051 milliequivalents + 0.005 milliequivalents + 0.0015 milliequivalents = 0.0116 milliequivalents.
Therefore, each 5 mL of the solution contains 0.0116 milliequivalents of potassium (K+).
To calculate the milliequivalents of potassium (K+) in each 5 mL of the solution, we need to determine the number of moles of potassium ions (K+) in each compound and then convert that to milliequivalents.
First, let's calculate the number of moles of potassium (K+) in each compound:
For potassium acetate (C2H3KO2):
Molar mass of K = 39.10 g/mol
Number of moles of K = 0.5 g / 39.10 g/mol
For potassium bicarbonate (KHCO3):
Molar mass of K = 39.10 g/mol
Number of moles of K = 0.5 g / 39.10 g/mol
For potassium citrate (C6H5K3O7 . H2O):
Molar mass of K = 39.10 g/mol
Number of moles of K = 0.5 g / 39.10 g/mol
Next, let's calculate the total number of moles of potassium (K+) in the 5 mL solution:
Total moles of K = moles of K in potassium acetate + moles of K in potassium bicarbonate + moles of K in potassium citrate
Finally, let's convert the total moles of potassium (K+) to milliequivalents:
Milliequivalents of K = Total moles of K × molar equivalent weight of K (1 milliequivalent = 1/1000th of an equivalent)
Equivalent weight of K = molar mass of K / number of equivalents of K (for each K ion, there is 1 equivalent)
Substituting the values, we can calculate the milliequivalents of K