Calculate the weight of solid NaOH required to prepare (a) 5 liters of a 2 M solution, (b) 2 liters of a solution of pH 11.5?
(a) How many mols do you want? That's mols = M x L = 2 x 5 = 10
Then grams = mols NaOH x molar mass NaOH = ?
(b) pH of 11.5 is a pOH of 2.5 (that's 14-11.5 = 2.5).
So (OH^-) in mols/L = -log(OH^-) = -log 2.5
OH^- = ? remember in mols/L. You want 2 L so mols you want will be M x L = ?
Then follow the first solution in a.
To calculate the weight of solid NaOH required for each scenario, we need to use some basic concepts of chemistry. Let's break down each question.
(a) 5 liters of a 2 M solution:
To calculate the weight of solid NaOH required to prepare this solution, we need to use the formula:
Molarity (M) = moles of solute / liters of solution
First, rearrange the formula to calculate moles of solute:
moles of solute = Molarity (M) * liters of solution
In this case, we have a 2 M solution and 5 liters of solution. Plugging these values into the formula:
moles of solute = 2 M * 5 L
moles of solute = 10 moles
Now, to calculate the weight of solid NaOH, we need to use its molar mass. The molar mass of NaOH is the sum of the individual atomic masses of sodium (Na), oxygen (O), and hydrogen (H).
Na = 22.99 g/mol
O = 16.00 g/mol
H = 1.01 g/mol
Thus, the molar mass of NaOH = 22.99 + 16.00 + 1.01 = 40.00 g/mol
Finally, to calculate the weight of solid NaOH:
weight of solid NaOH = moles of solute * molar mass
weight of solid NaOH = 10 moles * 40.00 g/mol
weight of solid NaOH = 400 g
Therefore, to prepare 5 liters of a 2 M solution, you would need 400 grams of solid NaOH.
(b) 2 liters of a solution of pH 11.5:
The pH of a solution provides information about its acid or base concentration. However, it does not directly provide information about the molarity or weight of the solute. Therefore, we need additional information to determine the weight of solid NaOH required.
Assuming the solution is a sodium hydroxide (NaOH) solution with a pH of 11.5, we need to calculate the molarity of the solution (M).
To convert the pH into molarity, we use the equation:
pH = -log[H+]
where [H+] represents the concentration of hydrogen ions in moles per liter.
First, calculate the concentration of hydrogen ions:
[H+] = 10^(-pH)
[H+] = 10^(-11.5) (since pH = 11.5 in this case)
[H+] = 3.16 x 10^(-12) moles/liter
As NaOH is a strong base, it dissociates into Na+ and OH- ions in solution. Since the ratio of Na+ to OH- ions in NaOH is 1:1, the concentration of OH- ions is also 3.16 x 10^(-12) moles/liter.
Now, we have the concentration of the OH- ions in moles/liter. To determine the molarity (M), we need to consider the volume of the solution.
In this case, we have 2 liters of solution. Therefore:
moles of solute = concentration of OH- ions * volume of solution
moles of solute = (3.16 x 10^(-12) moles/L) * 2 L
moles of solute = 6.32 x 10^(-12) moles
Finally, to calculate the weight of solid NaOH, we can follow the same steps as in scenario (a):
weight of solid NaOH = moles of solute * molar mass
weight of solid NaOH = 6.32 x 10^(-12) moles * 40.00 g/mol
weight of solid NaOH ≈ 2.53 x 10^(-10) grams
Therefore, to prepare 2 liters of a solution with a pH of 11.5, you would need approximately 2.53 x 10^(-10) grams of solid NaOH.