A student added solid K2O to a 750.0 mL volumetric flask. The solid was dissolved in water and then the flask filled with water to the mark. This formed 750.0 mL of the initial KOH solution. 20.0 mL of the initial solution was transferred to another flask and diluted to 200.0 mL. The pH of the diluted solution was 14.22.
(a) What is the concentration of hydroxide ion is the diluted solution?
M ***found this to be 1.66M
(b) What is the concentration of hydroxide ion in the original solution?
M
(c) What mass of K2O was added to the first flask?
not sure how to do the last two parts....
Show your work for part a and we may be able to help.
To calculate the concentration of hydroxide ion in the diluted solution, we can use the pH value given. The pH scale ranges from 0 to 14, where a pH of 7 is considered neutral, pH less than 7 is acidic, and pH greater than 7 is basic. A pH of 14.00 is equivalent to a hydroxide ion concentration of 1.0 M (Molar).
To calculate the concentration of hydroxide ion in the diluted solution (part a), we will use the formula:
pOH = 14.00 - pH
pOH is the negative logarithm of the hydroxide ion concentration. By solving for pOH, we can find the concentration of hydroxide ion in the diluted solution.
pOH = 14.00 - 14.22 = -0.22
Now, we can use the relationship between pOH and hydroxide ion concentration to find the concentration of hydroxide ion (OH-) in the diluted solution:
[H+] = 10^(-pOH)
[H+] = 10^(-(-0.22)) = 1.66 M
So, the concentration of hydroxide ion in the diluted solution is 1.66 M.
Moving on to part b, to find the concentration of hydroxide ion in the original solution, we can use the concept of dilution. Since 20.0 mL of the initial solution was transferred to another flask and diluted to 200.0 mL, we can use the formula of dilution:
C1V1 = C2V2
C1 is the initial concentration, V1 is the initial volume, C2 is the final concentration, and V2 is the final volume.
We know that C2 (final concentration) is 1.66 M (from part a), V1 (initial volume) is 20.0 mL, and V2 (final volume) is 200.0 mL.
By rearranging the formula, we can find C1 (initial concentration):
C1 = (C2V2) / V1
C1 = (1.66 M × 200.0 mL) / 20.0 mL
C1 = 16.6 M
Therefore, the concentration of hydroxide ion in the original solution is 16.6 M.
Moving on to part c, we need to determine the mass of K2O added to the first flask. To do this, we need to know the molar mass of K2O, which consists of potassium (K) and oxygen (O).
The molar mass of potassium (K) is 39.1 g/mol, and the molar mass of oxygen (O) is 16.0 g/mol. Since K2O has two atoms of potassium and one atom of oxygen, the molar mass of K2O is calculated as follows:
Molar mass of K2O = (2 × molar mass of K) + (1 × molar mass of O)
Molar mass of K2O = (2 × 39.1 g/mol) + (1 × 16.0 g/mol)
= 78.2 g/mol + 16.0 g/mol
= 94.2 g/mol
Now, to find the mass of K2O added to the first flask, we need to use the formula:
Mass = concentration × volume × molar mass
The volume is given as 750.0 mL in the problem, and the concentration of the initial solution (from part b) is 16.6 M. Substituting these values, we get:
Mass = 16.6 M × 750.0 mL × (1 L/1000 mL) × 94.2 g/mol
Mass = 9.3 g
Therefore, the mass of K2O added to the first flask is 9.3 grams.