Question

An antacid tablet weighing $1{\text{ g}}$ containing aluminium hydroxide as the only basic substance and the rest of its components being neutral, was dissolved in $200{\text{ mL}}$ of $0.1{\text{ M HCl}}$ . The excess ${\text{HCl}}$ was back titrated and required $90{\text{ mL}}$ of $0.1{\text{ N}}$ base for exact neutralisation. Milli equivalents of aluminium hydroxide in the sample of antacid tablet is:
(A) 9
(B) 11
(C) 12
(D) 20

Answer 1

Hint: Antacids are usually basic substances consumed at times of acidity in the stomach. In this question, an antacid made of aluminium hydroxide is used. The equivalents of aluminium hydroxide present in the antacid tablet have to be calculated using the back-titration method.

Complete step by step solution:
We are given an antacid tablet containing aluminium hydroxide as the basic substance was used and we are asked to calculate the milli equivalents of aluminium hydroxide present in a tablet. The given information are as follows:
Mass of antacid = $1{\text{ g}}$
The tablet is dissolved in a dilute solution of ${\text{HCl}}$ to neutralise the acid of known volume and concentration.
Volume of acid used = $200{\text{ mL}}$= $0.2{\text{ L}}$
Molarity of ${\text{HCl}}$ = $0.1{\text{ M}}$
No. of moles of acid used can be calculated using the following equation:
${\text{Molarity}} = \dfrac{{{\text{No}}{\text{. of moles}}}}{{{\text{Volume (in L)}}}}$ $\therefore $${\text{No}}{\text{. of moles = Molarity }} \times {\text{Volume (in L) }}$
 So, ${\text{No}}{\text{. of moles = 0}}{\text{.2 moles}}$ of ${{\text{H}}^ + }$

So, the whole of the base in the antacid tablet would have reacted with the excess of acid in which the antacid tablet that was dissolved. The excess of acid or unreacted ${{\text{H}}^ + }$present in the solution is quantitatively measured by back titrating it with a known concentration of base whose concentration is given.
Normality of base = $0.1{\text{ N}}$
Suppose that the normality and molarity of the base is the same.
Molarity of base = $0.1{\text{ M}}$
Volume of base used = $90{\text{ mL}}$= $0.09{\text{ L}}$
$\therefore $ ${\text{No}}{\text{. of moles of base used = Molarity }} \times {\text{Volume (in L) }}$
                                                        $ = {\text{ 0}}{\text{.1 mol}}{{\text{L}}^{ - 1}}{\text{ }} \times {\text{ 0}}{\text{.09 L}}$
                                                        $ = {\text{ 0}}{\text{.009 moles}}$

$0.009{\text{ moles}}$ of base is used to neutralise excess ${{\text{H}}^ + }$ present in the solution.

$\therefore $No. of moles of ${\text{HCl}}$ neutralised by antacid = Total no. of moles of ${{\text{H}}^ + }$- No. of moles of base used in back titration
                                                                                    $ = {\text{ 0}}{\text{.02 - 0}}{\text{.009 }}$
                                                                                    $ = {\text{ }}0.011{\text{ moles}}$
                                                                                    $ = {\text{ 11 mmol}}$

Therefore, one antacid tablet contains $11{\text{ milli moles}}$ or $11{\text{ milli equivalents}}$ of aluminium hydroxide in it.

Hence, option (B) is the correct answer.

Note:
Titration is a commonly used laboratory method for quantitative analysis to determine the unknown concentration of a substance called analyte by using known volume and concentration of another substance called the titrant by the chemical reaction between the titrant and the analyte.