How many drops of 35% do I need for ....

Question

I use 12 drops of 3% H2O2 in 1 teaspoon. I want to replace the 3% with 35% but keep it at the potency level of what 3% would offer in 1 teaspoon. How many drops of 35% would I need to do that?

I would like a explanation too on how you've figured it out. I'm not so knowledgeable in math but I learn quick & enjoy it.

[This is probably also very easy for most of you, sorry if it sounds stupid I haven't really received proper education in this field.]

Thanks.

Answer 1

Note that the number of drops of pure $H_2O_2$ in $n$ drops of $x \%$ concentrated $H_2O_2$ is given by $$ [\text{Amount }H_2O_2] = \frac{x}{100} \cdot n $$ So, for example, $4$ drops of $50\%$ $H_2 O_2$ is the same as $\frac{50}{100} \cdot 4 = 2$ drops of pure $H_2 O_2$. If we want to keep the same potency level, our goal is to keep the amount of pure $H_2O_2$ in a teaspoon exactly the same.

So, if $n$ is the number of drops of $35 \%$ that we need, then we know that $$ \frac{35}{100} \cdot n = \frac{3}{100} \cdot 12 $$ Use this information to solve for n.

Answer 2

What exactly you mean by potency? I assume that it is the product of no. of drops and the percentage concentration. Therefore, let 'x' drops of 35% H2O2 be equal to 12.3% potency. This gives the equation x.35=12.3 giving 'x' =36/35=1.02 drops.