Calculate the equivalent mass of $\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}$ when it is titrated against HCl in the presence of phenolphthalein.
A. 106
B. 53
C. 26.5
D. 212
Answer
612k+ views
Hint: We can say that the equivalent mass of compound is the total number of gram equivalents of the compound. Equivalent mass can be calculated by finding the ratio of molar mass of the substance to the acidity or basicity or n-factor of the same substance. Here sodium bicarbonate is titrated with HCl. So, we will have to consider the n-factor of the sodium bicarbonate. From the formula, we can clearly say that 1 molecule of carbonate will react with 2 molecules of sodium to form the sodium bicarbonate.
Complete step by step answer:
Sodium carbonate is titrated with hydrochloric acid to know the concentration of sodium carbonate with the help of a known concentration of hydrochloric acid.
It means that the fixed mass of an element can displace 1 part by mass of hydrogen, 35.5 parts by mass of chlorine and 8 parts by mass oxygen.
To calculate the equivalent mass of sodium carbonate firstly we will write the chemical equation:
\[\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}\text{ + HCl }\to \text{ NaCl + NaHC}{{\text{O}}_{3}}\]
Now, we will calculate the molar mass of sodium carbonate:
$\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}$
= $\left( \text{23 }\times \text{ 2} \right)\text{ + 12 + }\left( \text{16 }\times \text{ 3} \right)$
= $\text{46 + 12 + 48}$
= $\text{106}$
The formula to find the equivalent mass of sodium carbonate is:
Equivalent mass = $\dfrac{{Molar\;mass}}{{Number\; of\;hydrogens\;added\;or\;removed}}$
$\text{= }\dfrac{106}{1}\text{ }=\text{ 106}$
So, the equivalent mass of sodium carbonate is 106.
Therefore, option A is the correct answer.
Note: Equivalent mass has great use in the volumetric analysis because those compounds who have more equivalent mass are preferred over those compounds who have less equivalent mass. After all, the chance of error in weighing is very much less.
Complete step by step answer:
Sodium carbonate is titrated with hydrochloric acid to know the concentration of sodium carbonate with the help of a known concentration of hydrochloric acid.
It means that the fixed mass of an element can displace 1 part by mass of hydrogen, 35.5 parts by mass of chlorine and 8 parts by mass oxygen.
To calculate the equivalent mass of sodium carbonate firstly we will write the chemical equation:
\[\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}\text{ + HCl }\to \text{ NaCl + NaHC}{{\text{O}}_{3}}\]
Now, we will calculate the molar mass of sodium carbonate:
$\text{N}{{\text{a}}_{2}}\text{C}{{\text{O}}_{3}}$
= $\left( \text{23 }\times \text{ 2} \right)\text{ + 12 + }\left( \text{16 }\times \text{ 3} \right)$
= $\text{46 + 12 + 48}$
= $\text{106}$
The formula to find the equivalent mass of sodium carbonate is:
Equivalent mass = $\dfrac{{Molar\;mass}}{{Number\; of\;hydrogens\;added\;or\;removed}}$
$\text{= }\dfrac{106}{1}\text{ }=\text{ 106}$
So, the equivalent mass of sodium carbonate is 106.
Therefore, option A is the correct answer.
Note: Equivalent mass has great use in the volumetric analysis because those compounds who have more equivalent mass are preferred over those compounds who have less equivalent mass. After all, the chance of error in weighing is very much less.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

