One 240. mL serving of ChemCola™ contains 57.0 mg phosphorus (which is found in the phosphoric acid, H3PO4, in ChemCola). What volume (in mL) of 0.0950 M NaOH is needed to neutralize the H3PO4 in one serving of ChemCola™?
H3PO4 + 3NaOH ==> Na3PO4 + 3H2O
How many mols in 57 mg of P. Mols = grams/atomic mass = 0.057/approx 31 but you need to redo that and all calculations that follow. I've just estimated all calculations. mols P = 0.00184 mols. That's also 0.00184 mols H3PO4. Now convert mols H3PO4 to mols NaOH.
That's 0.00184 mols H3PO4 x (3 mols NaOH/1 mol H3PO4) = estimated 0.0055 mols NaOH.
Then mols NaOH = M x L = 0.0055 = M x L. You know M from the problem of the NaOH, substitute and solve for L, then convert to mL. Post your work if you get stuck.
Well, let's break this down! We know that the molarity (M) of NaOH is 0.0950, and we want to find the volume of NaOH needed to neutralize the phosphoric acid (H3PO4).
To figure out the volume, we need to use the equation:
M1V1 = M2V2
The M1 in this equation is the molarity of H3PO4, which we don't have. But lucky for us, we have the number of milligrams of phosphorus in one serving of ChemCola!
So, let's convert the 57.0 mg of phosphorus to moles. The molar mass of phosphorus is approximately 31.0 g/mol. So, 57.0 mg is equal to 57.0/31.0 = 1.84 mmol (rounded to two decimal places).
Now, let's look at the balanced equation for the neutralization reaction between H3PO4 and NaOH:
H3PO4 + 3NaOH -> Na3PO4 + 3H2O
From the equation, we can see that one mole of H3PO4 reacts with 3 moles of NaOH.
Since the ratio of moles is 1:3, we can say that the number of moles of NaOH needed is 3 times the number of moles of H3PO4, which in this case is 3 times 1.84 mmol = 5.52 mmol.
Now that we have the number of moles of NaOH, we can use the molarity to find the volume of NaOH required:
0.0950 M x V2 = 5.52 mmol
Now, we just need to solve for V2:
V2 = 5.52 mmol / 0.0950 M
V2 ≈ 58.1 mL
So, approximately 58.1 mL of 0.0950 M NaOH is needed to neutralize the H3PO4 in one serving of ChemCola.
To determine the volume of 0.0950 M NaOH needed to neutralize the H3PO4 in one serving of ChemCola™, we need to use stoichiometry.
The balanced chemical equation for the neutralization of H3PO4 with NaOH is:
H3PO4 + 3NaOH -> Na3PO4 + 3H2O
From the equation, we can see that 1 mole of H3PO4 reacts with 3 moles of NaOH.
To find the moles of H3PO4 in one serving of ChemCola™, we convert the given mass of phosphorus (57.0 mg) to moles using the molar mass of phosphorus (P), which is 30.97 g/mol.
Moles of phosphorus = (57.0 mg) / (30.97 g/mol) = 0.00184 mol
Since the stoichiometric ratio is 1:3, the moles of NaOH needed for neutralization is:
Moles of NaOH = 3 * Moles of H3PO4 = 3 * 0.00184 mol = 0.00552 mol
Now, we can use the molarity and moles of NaOH to calculate the volume needed.
Volume of NaOH = Moles of NaOH / Molarity of NaOH
Volume of NaOH = 0.00552 mol / 0.0950 mol/L = 0.0581 L
Finally, we convert the volume from liters to milliliters:
Volume of NaOH = 0.0581 L * (1000 mL / 1 L) = 58.1 mL
Therefore, approximately 58.1 mL of 0.0950 M NaOH is needed to neutralize the H3PO4 in one serving (240 mL) of ChemCola™.
To find the volume of 0.0950 M NaOH needed to neutralize the H3PO4 in one serving of ChemCola, we need to use stoichiometry and balance the chemical equation.
The balanced chemical equation for the reaction between phosphoric acid (H3PO4) and sodium hydroxide (NaOH) is:
H3PO4 + 3NaOH -> Na3PO4 + 3H2O
From the balanced equation, we can see that 1 mole of H3PO4 reacts with 3 moles of NaOH.
First, let's convert the amount of phosphorus given (57.0 mg) to moles. The molar mass of phosphorus (P) is approximately 31.0 g/mol.
57.0 mg phosphorus * (1 g / 1000 mg) * (1 mol / 31.0 g) = 0.001838 moles phosphorus
Since phosphoric acid (H3PO4) has a 1:1 stoichiometric ratio with phosphorus (P), we can determine the number of moles of H3PO4 present in one serving of ChemCola:
0.001838 moles H3PO4
To neutralize this amount of H3PO4, we need 3 times this number of moles of NaOH. So,
0.001838 moles H3PO4 * 3 moles NaOH / 1 mole H3PO4 = 0.005514 moles NaOH
Now, we can use the molarity and the number of moles of NaOH to find the volume (in liters) of 0.0950 M NaOH:
0.0950 moles NaOH / 1 L * (0.005514 moles NaOH / 1) = 0.000528 L NaOH
Finally, we can convert the volume from liters to milliliters:
0.000528 L NaOH * (1000 mL / 1 L) = 0.528 mL NaOH
Therefore, approximately 0.528 mL of 0.0950 M NaOH is needed to neutralize the H3PO4 in one serving of ChemCola™.