In the laboratory, $20.0$ milliliters of an aqueous solution of calcium hydroxide, $\ce{Ca(OH)2}$, was used in a titration. A drop of phenolphthalein was added to it to indicate the end point. The solution turned colorless after $20.0$ milliliters of a standard solution of $\pu{0.050 M}$ $\ce{HCl}$ solution was added. What was the molarity of the $\ce{Ca(OH)2}$?
(A) $\pu{0.010 M}$
(B) $\pu{0.025 M}$
(C) $\pu{0.50 M}$
(D) $\pu{0.75 M}$
(E) $\pu{1.0 M}$
I did not know how to answer such kind of question using formulas, that is why I applied logic. At first, I wrote such a reaction:
$$\ce{\underset{\pu{20 mL}}{Ca(OH)2} + \underset{\pu{0.05 M}}{\underset{\pu{20 mL}}{HCl}} = end point}$$
Then, I understood that if the volume of both reactants is equal, the molarity of $\ce{HCl}$ must be higher than that of $\ce{Ca(OH)2}$, as $\ce{HCl}$ changes the nature of $\ce{Ca(OH)2}$.
As I knew that the molarity of $\ce{Ca(OH)2}$ is just a bit less than that of $\ce{HCl}$, I chose B, and this variant turned out to be the right one.
In the book, I have such explanation:
In the titration, the reaction is:
$$\ce{2HCl + Ca(OH)2 = CaCl2 + 2H2O}$$ The acid to base ratio is $2 : 1$, or (moles acid used) = 2(moles base used), so $M_\mathrm{a}V_\mathrm{a} = 2M_\mathrm{b}V_\mathrm{b}$, where $M$ is the molarity and $V$ is the volume expressed in liters. Then
$$M_\mathrm{b} = \frac{M_\mathrm{a}V_\mathrm{a}}{2V_\mathrm{b}}$$ $$M_\mathrm{b} = \frac{\pu{0.05 M} \times \pu{0.02 L}}{2}$$ $$\pu{0.02 L} = \pu{0.025 M}$$
I do not understand the explanation. Although it comes up with the same variant as I did, if I had such formula I would use it this way $2M_\mathrm{a}V_\mathrm{a} = M_\mathrm{b}V_\mathrm{b}$ as we have 2 moles of acid.
Then, the answer would be $0.1$. The same formula is used on this forum. Can you help me understand what is wrong here please?