Name: Balancing Redox Reactions with Oxidation Numbers Method
Uploaded: 2019-07-19
Duration: 8 min 51 s
Description: The oxidation number is the total number of charges on an ion formed by the gain or loss of electrons. How do we balance redox reactions by using them? Learn...

Question

Acid is generated in the stomach when the concentration of acid goes up, a man suffers from acidity. The doctor prescribes him antacid. By taking

3

antacid tablets, the strength decreases from

(\frac{N}{10})

to

(\frac{N}{200})

. If the fluid material is

100 ml

, calculate the amount of base per antacid present per antacid tablet. (Equivalent mass of base in antacid

= 100

).

Accepted Answer

Normality is defined as the number of gram equivalents of solute, present in one litre of solution.

Given that, normality of acid before reaction is

$N_{1} = \frac{1}{10}$ .

Volume of acid given is $V = 100 ml = 0.1 l$

Let the number of moles of acid be $m_{1}$ and the equivalent weight of acid be $n_{1}$ . We know that,

$N_{1} \times V_{1} = no . of moles \times equivalent number \frac{1}{10} \times 0.1 = m_{1} \times n_{1} m_{1} \times n_{1} = 0.01 m_{1} = \frac{1}{100 \times n_{1}}$

When the acid reacts with antacid, let the number of moles of acid be $m_{2}$ .

Normality of acid after it reacts with antacid is $N_{2} = \frac{1}{200}$ .

Volume of acid is $100 ml$ . We know that,

$N_{2} \times V_{2} = no . of moles of acid after reaction \times equivalent number \frac{1}{200} \times 0.1 = m_{2} \times n_{2} m_{2} \times n_{2} = 0.002 m_{2} = \frac{1}{2000 \times n_{2}}$

The actual number of moles that take part in the reaction is given by,

$m = m_{1} - m_{2} m = \frac{1}{100 n} - \frac{1}{2000 n} m = \frac{19}{2000 n}$

Let the equivalent number of base be $n_{o}$ . GIven number of tablets are $3$ . We know that,

$No . of tablets \times no . of moles of base per tablet \times n_{o} = eq . number of acid \times no . of moles of acid reacted 3 \times \frac{mass of base per tablet}{molar mass of base} \times n_{o} = \frac{19}{2000 n} \times n 3 \times \frac{mass of base per tablet}{\frac{molar mass of base}{n_{o}}} = \frac{19}{2000} 3 \times \frac{mass of base per tablet}{equivalent mass of base} = \frac{19}{2000} 3 \times mass of base per tablet = \frac{19}{2000} \times 100 mass of base per tablet = \frac{19}{60} = 0.317$

Important Questions on Redox Reaction

Learn and Prepare for any exam you want !