# Chemistry

A solution household bleach contains 5.25% NaClO, by mass. Assuming that the density of bleach is the same as water(1.0g/ml). Calculate the volume in mL of household bleach that should be diluted with water to make 500.0 ml of a pH= 9.80 solution.

Here's what I've done so far..
Kw = Ka * Kb
1.0 x 10^-14 = 4.0 x 10^-8 * Kb
Kb = 2.5 x 10^-7

5.25% = 52.5 g NaClO/74.44 g.mol = 0.705 mol NaClO

where would i go from here?

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1. I would do this.
1.00 g/mL x 1000 mL x 0.0525 x (1/molarmass NaClO) = about 0.7 M for the 5.25% bleach. That's what we have. What do we want?

If pH = 9.80, then pOH = 4.20 and OH^- = about 6.3E-5M
..........ClO^- + HOH ==> HClO + OH^-
Equil....x.......x.....6.3E-5..6.3E-5
(Note: You should go through and clean this up).
Kb from your work above = (HClO)(OH^-)/(ClO^-)
Substitute from the chart above and solve for x = (ClO^-), then use
c1v1 = c2v2
c = concn
v = volume
You know c1(about 0.7M),you want to find v1 (in mL), c2 = x from above and v2 = 500 mL. Solve for v1.

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2. Thanks Dr. Bob! This really helped.

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