What mass of KOH is necessary to prepare 799.9 mL of a solution having a pH = 11.68?
To determine the mass of KOH necessary to prepare a solution with a pH of 11.68, we need to consider the balanced chemical equation for the ionization of KOH in water:
KOH(aq) → K+(aq) + OH-(aq)
Since KOH is a strong base, it will fully dissociate in water and provide one hydroxide ion (OH-) for every molecule of KOH. The concentration of hydroxide ions (OH-) can be used to calculate the concentration of KOH in the solution.
First, let's convert the pH to the concentration of hydroxide ions (OH-) using the equation:
pOH = 14 - pH
pOH = 14 - 11.68
pOH = 2.32
Next, we can convert the pOH to the concentration of KOH using the equation:
[OH-] = 10^(-pOH)
[OH-] = 10^(-2.32)
[OH-] ≈ 0.005746 M
Now, let's calculate the number of moles of KOH (n) required using the equation:
n = [OH-] * volume
n = 0.005746 M * 0.7999 L
n ≈ 0.004595 mol
The molar mass of KOH is 39.10 g/mol (potassium: 39.10 g/mol, oxygen: 16.00 g/mol, hydrogen: 1.01 g/mol).
Finally, we can calculate the mass of KOH (m) using the equation:
m = n * molar mass
m = 0.004595 mol * 56.106 g/mol (39.10 g/mol + 16.00 g/mol + 1.01 g/mol)
m ≈ 0.257 g
Therefore, approximately 0.257 grams of KOH is necessary to prepare 799.9 mL of a solution with a pH of 11.68.
To determine the mass of KOH needed to prepare the solution, we need to follow these steps:
Step 1: Understand the problem:
We are given the volume of the solution (799.9 mL) and the pH (11.68). We want to find the mass of KOH needed to prepare this solution.
Step 2: Recall the definition of pH:
pH is a measure of the acidity or basicity of a solution. It is defined as the negative logarithm (base 10) of the concentration of hydrogen ions (H+) in the solution.
Mathematically, pH = -log[H+]
Since we are given the pH (11.68), we can find the concentration of hydroxide ions (OH-) in the solution. Recall that in a basic solution, the concentration of OH- is related to the concentration of H+ by the equation:
[H+] × [OH-] = 1.0 × 10^(-14) at 25°C
Step 3: Calculate the concentration of OH- ions:
To find the OH- concentration, we need to use the above equation and the pH value.
[H+] = 10^(-pH)
[H+] = 10^(-11.68)
[OH-] = 1.0 × 10^(-14) / [H+]
Now we can substitute the value of [H+] to calculate [OH-].
Step 4: Calculate the amount of KOH:
To calculate the amount of KOH needed, we need to use the concentration of OH- (in mol/L) and the volume of the solution (in L). The equation is:
Amount (in moles) = Concentration (in mol/L) × Volume (in L)
However, before we can proceed, we need to convert the volume from milliliters to liters:
799.9 mL = 799.9 mL × (1 L / 1000 mL)
Now, we have the volume in liters and all the information required to calculate the amount of KOH.
Step 5: Convert the amount to mass:
To convert the amount of KOH from moles to grams, we need to use its molar mass. The molar mass of KOH is the sum of the atomic masses of its constituent elements: potassium (K), oxygen (O), and hydrogen (H).
Now let's go through the calculations step by step.
Convert pH to pOH.
pH + pOH = pKw = 14.
Then convert pOH to OH^-
That will be the molarity of KOH.
M = moles/L solution.
Solve for moles, then
moles = grams/molar mass
Solve for grams.