Question

How many grams of $Mg{(OH)_2}$ will be needed to neutralize 25 mL of stomach acid if stomach acid is 0.10 M $HCl$ ?

Answer 1

Hint: The compound can broadly be classified either as neutral, as the base of an acid. The acidic strength is related to the production of hydroxyl ions while the strength of the base is determined by the hydronium ion concentration. These are the ions that are formed when the ions interact with water molecules. When bases and acids react with each other they tend to neutralize each other.

Complete step by step answer:
The reaction that will happen upon the reaction of the compounds in the question can be represented as follows
$2HCl + Mg{(OH)_2} \to MgC{l_2} + 2{H_2}O$
The volume of the acid given is 5 mL as specified in the question. The molarity of the acid is also specified in the question as 0.10 M
The molarity of the solution can be represented in terms of the volume of the solution rather than the mass of the solvent as in the case of molality. It can be represented as
${{Molarity (M) = }}\dfrac{{{{Moles\; of \;solute}}}}{{{{Volume\; of \;solution\; in\; litre}}}}$
.From the formula of calculating molarity we can deduce the number of moles the acid contains. This can be done as
$Moles\,of\,HCl = 0.10 \times 0.025L = 0.0025$
From the stoichiometric coefficient of the compound in the above reaction, we conclude that the number of moles of magnesium hydroxide will be half of the number of moles of acid thus,
$Moles\,of\,Mg{(OH)_2} = \dfrac{{0.0025}}{2} = 0.00125$
Since the moles of the compound can be calculated from the equation given as
$Moles = \dfrac{{mass}}{{molar\;mass}}$

So, $Mass\,of\,Mg{(OH)_2} = 0.00125 \times 58.327 = 0.0729g$.

Note: There are other representations of the concentration of solution than molarity.
While molality is represented by $m$ another representation, Molarity is represented by $M$ .
Molarity of the solution can be represented in terms of the volume of the solution rather than the mass of the solvent as in the case of molality
For the quantitative representation of the concentration of the solution several ways can be used, molality is one of those criteria.
Molality is represented by m and can be calculated by the following formula
$Molality(m) = \dfrac{{{{Moles\; of \;solute}}}}{{{{Mass\; of\; solvent\; in\;kg}}}}$