Name: Understanding Redox Titrations
Uploaded: 2019-07-19
Duration: 11 min 41 s
Description: Redox titration is a great way to identify whether a reaction is completed or not, where we know that reaction is complete, and it helps us find the strength...

Question

20 {cm}^{3}

of

1 M

Sulphuric acid was neutralised by

25 {cm}^{3}

of ammonia solution. Calculate the concentration of ammonia solution.

Accepted Answer

Given:

$20 {cm}^{3}$ of $1 M$ Sulphuric acid ( $H_{2} {SO}_{4}$ )

$25 {cm}^{3}$ of ammonia ( ${NH}_{3}$ ).

Concentration of ammonia $({NH}_{3})$ = ?

Now the reaction for the above equation will be :

$2 {NH}_{3} + H_{2} {SO}_{4} \to {(NH 4)}_{2} {SO}_{4}$

We know that : $Molarity = \frac{(number of moles)}{(Volume in L)}$

Therefore, $Number of mole = \frac{(Molarity x Volume)}{1000} = \frac{1 x 20}{1000} = 0.02 mole$

$0.02$ mole of $H_{2} {SO}_{4}$ is present in the solution.

The molar ratio from the above equation we have: $2 : 1$

It means that $1$ mole of $H_{2} {SO}_{4}$ requires $2$ mole of ${NH}_{3}$ .

Therefore, Total number of moles of ${NH}_{3} = 0.02 x 2 = 0.04$ mole.

Now we can calculate molarity of ${NH}_{3}$ , $Molarity = \frac{(Number of mole x 1000)}{Volume} = \frac{0.04 x 1000}{25} = 1.6 M$

$20 {cm}^{3}$ of $1 M$ Sulphuric acid was neutralised by $25 {cm}^{3}$ of ammonia solution. Calculate the concentration of ammonia solution.

Important Questions on Acids and Bases

Learn and Prepare for any exam you want !

20cm3 of 1M Sulphuric acid was neutralised by 25cm3of ammonia solution. Calculate the concentration of ammonia solution.

Important Questions on Acids and Bases

Learn and Prepare for any exam you want !

$20 {cm}^{3}$ of $1 M$ Sulphuric acid was neutralised by $25 {cm}^{3}$ of ammonia solution. Calculate the concentration of ammonia solution.