A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mass. Assuming that the density of bleach is the same as water, calculate the volume of household bleach that should be diluted with water to make 500.0 mL of a pH = 10.10 solution.

If you look into your inner self, you find the answer if pi

Dont listen to that answer, its wrong

Sean is right

Yall are the worst

Thanks for the help, Sean!! Perfect answer

To calculate the volume of household bleach needed to make a pH = 10.10 solution, we need to determine the amount of sodium hypochlorite (NaOCl) present in the desired final solution.

Step 1: Calculate the moles of NaOCl needed:
To do this, we need to convert the pH of the solution to a concentration of hydroxide ions (OH-). This is because in an alkaline solution like this, the hydroxide ions determine the pH.

pOH = 14 - pH
pOH = 14 - 10.10
pOH = 3.90

Now, we can convert pOH to the concentration of hydroxide ions using the equation:

[OH-] = 10^(-pOH)

[OH-] = 10^(-3.90)
[OH-] = 7.94 x 10^(-4) M

Since the concentration of hydroxide ions is equal to the concentration of sodium hypochlorite in this case, we have:

[NaOCl] = 7.94 x 10^(-4) M

Step 2: Calculate the moles of NaOCl based on desired final volume:
We know that the desired volume of the final solution is 500.0 mL, which is equal to 0.500 L.

Moles of NaOCl = [NaOCl] x V
Moles of NaOCl = (7.94 x 10^(-4) M) x (0.500 L)
Moles of NaOCl = 3.97 x 10^(-4) mol

Step 3: Calculate the mass of NaOCl in the moles that we calculated:
To do this, we need to use the molar mass of NaOCl.

The molar mass of NaOCl = (sodium atomic mass) + (oxygen atomic mass) + (chlorine atomic mass)
= (23 g/mol) + (16 g/mol) + (35.5 g/mol)
= 74.5 g/mol

Mass of NaOCl = Moles of NaOCl x Molar mass of NaOCl
Mass of NaOCl = (3.97 x 10^(-4) mol) x (74.5 g/mol)
Mass of NaOCl = 0.0296 g

Step 4: Calculate the volume of household bleach:
The solution of household bleach contains 5.25% sodium hypochlorite by mass. This means that in 100 g of bleach, there are 5.25 g of sodium hypochlorite.

To calculate the volume, we can use the ratio of the mass of sodium hypochlorite to the density of water:

Volume of bleach = (Mass of NaOCl / % NaOCl in bleach) / Density of water
Volume of bleach = (0.0296 g / 0.0525) / 1 g/mL
Volume of bleach = 0.563 L or 563 mL

So, you would need to dilute 563 mL of household bleach with water to make 500 mL of a pH = 10.10 solution.

5.25% NaOCl by mass means 5.25 g NaOCl/100 g soln and if the density is 1.00 g/mL (as is water) then molarity NaOCl solution is (5.25/74.44) mols NaOCl in 0.1L; therefore, M = 0.075/0.1L = estimated 0.71M. For pH = 10.10 is pOH 14-10.10 = 3.90 and since pOH = -log(OH^-) then (OH^-) = about 1.3E-4 but you need to do that more accurately.

OCl^- hydrolyzes in H2O as
........OCl^- + HOH ==> HOCl + OH^-
I........x..............0......0
C...-1.3E-4...........1.3E-4..1.3E-4
E....x-1.34E-4........1.3E-4..1.3E-4

Kb for OCl = (Kw/Ka for HOCl) = (1.3E-4)^2/(x-1.3E-4). You know Kw and Ka, solve for x = (OCl^-) needed. That comes out to approximately 0.05 but you should confirm that. I just did a quickie on the calculator.

Then c1v1 = c2v2
c = concn
v = volume
0.71*v1 = 0.05*500 mL
Solve for v1.