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Wiki and lots of sources all say the same thing. "1 part of 3% hydrogen peroxide to 10 parts water". But I am not trying to neutralize water I want to neutralize laundry bleach. It could mean add 1 part hydrogen peroxide to 10 parts water and then use that to neutralize your bleach product but without knowing the concentration of sodium hypochlorite that the mixture was intended for that would not make sense either. The directions are a complete mystery no matter how you look at them. So I will ask in the following manner to make it clear.

We have a gallon of 3% sodium hypochorite household bleach. How much 3% hydrogen peroxide should be added to neutralize it and what will the neutralized product be that remains? The chemical reaction formula would be nice but I might be asking to too much here.

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    $\begingroup$ Does this answer your question? How to calculate the volume of oxygen in a reaction between hydrogen peroxide and bleach using this apparatus $\endgroup$ Jun 19 at 17:09
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    $\begingroup$ Out of curiosity, is there any particular reason to not just use up laundry bleach by using it in the laundry? After all, white clothing dinginess, like rust, never sleeps. $\endgroup$
    – Ed V
    Jun 19 at 20:38
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    $\begingroup$ Never mix bleach under any circumstances or for any reason. If you want to get rid of it call your county's hazardous waste disposal department and they can give you directions of where you can take it for safe disposal. You can also call the manufacturer of the product and they can give you directions for disposing of it in the toilet if it is under a certain amount AND copious amounts of water are used with multiple flushes. They can give the specifics. $\endgroup$
    – Sedumjoy
    Jul 14 at 4:31
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    $\begingroup$ Never mix bleach with anything. As far as hydrogen peroxide goes the combination can explode from what I read. To dispose of it call your county hazardous waste department. $\endgroup$
    – Sedumjoy
    Jul 14 at 4:38
  • $\begingroup$ ......or use it in your laundry. $\endgroup$
    – Sedumjoy
    Jul 14 at 16:08
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Wiki and lots of sources all say the same thing. "1 part of 3% hydrogen peroxide to 10 parts water."

When talking about aqueous solutions, the concentrations can be expressed as percent by weights, $\% (w/w)$, or by volume, $\% (v/v)$, or by weight in parts of solute by weight in 100 part of solution by volume, $\% (w/v)$. The above statement in the question means if you take 100 part of the solution by weight (e.g., $\pu{100 g}$ of solution), it contains 3 part of hydrogen peroxide and 97 parts of water by weights $\left(3\% (w/w)\right)$. It also can mean that it may contain 3 part of hydrogen peroxide to 97 parts of water by volumes $\left(3\% (v/v)\right)$ if the lebel doesn't say it is whether $\% (w/w)$ or $\% (v/v)$. The third kind is 3 part of hydrogen peroxide by weight to 100 parts of solution by volume $\left(3\% (w/v)\right)$, which is common for prepared chemical reagents to use in the laboratories.

Commercial $\ce{H2O2}$ solution is usually in $\% (v/v)$. However, when prepared as a chemical reagent, it is $\% (w/v)$. That means $\pu{100 mL}$ of $3\% (w/v)$ $\ce{H2O2}$ solution has $\pu{3 g}$ or the molarity of the solution:

$$3\% (w/v) = \frac{\pu{3 g} \ \ce{H2O2}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 g} \ \ce{H2O2}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{30 g} \ \ce{H2O2}}{\pu{34.01 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{0.882 mol L-1}$$

On the other hand, since commercial $\ce{H2O2}$ solution is usually in $\% (v/v)$, and density of pure $\ce{H2O2}$ is $\pu{1.450 g mL-1}$ at $\pu{20 ^\circ C}$ (Wikipedia):

$$3\% (v/v) = \frac{\pu{3 mL} \ \ce{H2O2}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 mL}\cdot \pu{1.45 g mL-1} \ \ce{H2O2}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{43.5 g} \ \ce{H2O2}}{\pu{34.01 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{1.28 mol L-1}$$

Household bleach is aqueous $\ce{NaClO}$ $(\pu{74.44 g mol-1})$ solution, sometimes included certain percent of $\ce{NaOH}$. If you have $3\%$ solution, you can do the same calculations as above to find out the correct molarity if the solution is given whether $3\% (w/v)$ or $3\% (v/v)$ (and it is without $\ce{NaOH}$). Usually, it is always given in $\% (w/v)$. Sometimes, the default densities may be provided instead of percentages by the manufacturer, which may differ for the different manufacturer since the manufacturer may change the density by adding salt or other chemicals to the solution. Therefore, it is hard to calculate molarity using provided default density if you don't know what else other than $\ce{NaClO}$.

Let's consider the solution you have is $3\% (w/v)$ $\ce{NaClO}$:

$$3\% (w/v) = \frac{\pu{3 g} \ \ce{NaOCl}}{\pu{100 mL} \ \text{soln}} = \frac{\pu{30 g} \ \ce{NaOCl}}{\pu{1000 mL} \ \text{soln}} = \frac{\frac{\pu{30 g} \ \ce{NaOCl}}{\pu{74.44 g mol-1}}}{\pu{1.0 L} \ \text{soln}} = \pu{0.403 mol L-1}$$

The reaction is:

$$\ce{NaClO + H2O2 -> NaCl + H2O + O2}$$

Accordingly, $\pu{1.0 mol}$ of $\ce{NaOCl}$ will react with $\pu{1.0 mol}$ of $\ce{H2O2}$. Therefore, the volume $(V_\ce{H2O2})$ of $3\% (w/v)$ $\ce{H2O2}$ needed is react completely with $\pu{1.0 L}$ of $3\% (w/v)$ $\ce{NaOCl}$ can be calculated using $M_1V_1 = M_2V_2$ as follows:

$$M_\ce{H2O2}V_\ce{H2O2} = M_\ce{NaOCl}V_\ce{NaOCl} \ \Rightarrow \ V_\ce{H2O2} = \frac{M_\ce{NaOCl}V_\ce{NaOCl}}{M_\ce{H2O2}} \\ = \frac{\pu{0.403 mol L-1}\cdot \pu{1.0 L}}{\pu{0.882 mol L-1}} = \pu{0.457 L}$$

Therefore, since you have one US gallon of bleach, which equals to $\pu{3.785 L}$, you need $0.457 \times 3.785 = \pu{1.73 L}$ of $3\% (w/v)$ $\ce{H2O2}$ solution.

Your final solution is $\frac{0.403}{1.457} = \pu{0.277 mol L-1}$ $\ce{NaCl}$ solution (of course, only if manufacturers did not add additives).

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The equation is: $\ce{NaClO + H2O2 -> NaCl + H2O + O2}$

A $3\%$ sodium hypochlorite contains $\pu{30 g}$ $\ce{NaClO}$ $(\pu{0.40 mol})$ per liter. So $\pu{0.1 mol}$ $\ce{NaClO}$ is contained in $\pu{250 mL}$ bleach.

A $3\%$ hydrogen peroxide contains $\pu{30 g}$ $\ce{H2O2}$ $(\pu{0.88 mol})$ per liter. So $\pu{0.1 mol}$ $\ce{H2O2}$ is contained in $\pu{114 mL}$ of your hydrogen peroxide solution.

So mix $\pu{250 mL}$ bleach $+$ $\pu{114 mL}$ of your hydrogen peroxide solution, or any multiple or submultiple, like $1$ liter bleach and $\pu{456 mL}$ peroxide solution.

You'll obtain a dilute salt solution.

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