How many milliliters of a 15.0%, by mass solution of KOH(aq) ( d= 1.14g/mL ) are required to produce 27.0 L of a solution with pH = 11.70?
I'm really not sure how to do this, it's from Acids & Bases chapter.
Any help is appreciated, thanks. (:
Nvm. I got it.
It wasn't so hard after all :D
You want pH = 11.70; therefore, pOH = 14.0-11.70 = ??
Then find OH^- from pOH and (OH^-) = ?? M.
How many moles must you have if you want that molarity (OH%-). M x L = moles OH^-.
What is the molarity of the 15.0% KOH solution?
The mass of 1 L of solution is
1000 mL x 1.14 g/mL = yy grams.
How much of that is KOH?
yy x 0.15 = zz grams KOH.
How many moles KOH is that?
zz grams/molar mass KOH = mols KOH in 1 L which is that molarity.
Then MKOH soln x LKOH soln = moles
M KOH you have. L KOH you want. and you know how many moles you want in the 27.0 L. Find L KOH and convert to mL.
To solve this problem, we need to use the equation for the pH of a basic solution:
pOH = 14.00 - pH
Since pH of the solution is given as 11.70, we can find the pOH:
pOH = 14.00 - 11.70
pOH = 2.30
Next, we need to convert the pOH to the OH- concentration using the equation:
OH- concentration = 10^(-pOH)
OH- concentration = 10^(-2.30)
OH- concentration = 0.0050119 M
Since KOHa
To solve this problem, we need to follow a series of steps. Let's break it down:
Step 1: Calculate the number of moles of OH⁻ needed to achieve pH 11.70.
Knowing the pH of a solution, we can convert it to pOH using the equation pOH = 14 - pH. In this case, the pOH is equal to 14 - 11.70 = 2.30.
Now, we can convert pOH to OH⁻ concentration using the equation pOH = -log[OH⁻]. Taking the antilog of both sides, 10^-2.30 = [OH⁻].
The OH⁻ concentration is 10^-2.30 = 0.004466 moles/L.
Step 2: Calculate the number of moles of KOH needed to achieve the desired OH⁻ concentration.
Since KOH is a strong base, it dissociates completely in water to produce one mole of OH⁻ per mole of KOH. Therefore, the number of moles of KOH is also 0.004466 moles/L.
Step 3: Calculate the mass of KOH needed using the molar mass of KOH.
The molar mass of KOH is 39.1 g/mol (potassium) + 16.0 g/mol (oxygen) + 1.0 g/mol (hydrogen) = 56.1 g/mol.
Therefore, the mass of KOH needed is 0.004466 moles/L × 56.1 g/mol = 0.250 g/L.
Step 4: Calculate the volume of the 15.0% KOH solution needed.
To do this, we'll use the formula:
% by mass = (mass solute / mass solution) × 100%
Here, % by mass = 15.0% and mass solution = 0.250 g/L.
Let's assume the volume of the KOH solution needed is V mL. We can set up the equation as:
15.0% = (0.250 g / V mL) × 100%
Solving for V:
(V mL) = (0.250 g) / (15.0 g / 100) = 1.67 mL.
So, 1.67 mL of the 15.0%, by mass solution of KOH is required to produce 27.0 L of a solution with pH = 11.70.