Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# One lit. of aqueous solution of ${H_2}S{O_4}$ contains 4.9 $g$ of the acid. 100 $ml$ of this solution is taken in a 1 lit. flask and diluted with water up to the mark. The $pH$ of the dilute solution isA. 2.0B. 2.301C. 1.699D. 3.699

Last updated date: 14th Sep 2024
Total views: 420.3k
Views today: 9.20k
Verified
420.3k+ views
Hint: When the volume of a sample is increased by just the addition of water then the process is termed as dilution. The dilution not only increases the volume of the sample but decreases the concentration of the sample too. Upon dilution since the concentration of the sample decreases it may start to show properties of the dilute version of the sample and that may affect its dissociation power as a whole.

$pH$ of a sample is the power of the hydrogen ions that the sample gives. The $pH$ of the samples ranges from 1 to 14, 1 being the most acidic and 14 the most basic.
The above question given the sample of ${H_2}S{O_4}$ which is an acid so it is most likely to have the $pH$ below 7.

${H_2}S{O_4}$ dissociates in the solution as

${H_2}S{O_4}\xrightarrow[{}]{}2{H^ + } + SO_4^{2 - }$

In order to find the $pH$ we need to calculate the concentration of ${H_2}S{O_4}$ in the solution which gives the hydrogen ion.

$Moles = \dfrac{\text{mass}}{\text{molar mass}}$,

Molar mass of ${H_2}S{O_4}$ can be calculated by adding the molar mass of its consisting atoms as

$2(Mass\,of\,H) + mass\,of\,S + 4(mass\,of\,O)$
$\Rightarrow 98$

Hence putting the values of mass and molar mass in the above mole relation we have

$moles = \dfrac{{4.9}}{{98}}$

$\Rightarrow moles = 0.05$

So, since the ${H_2}S{O_4}$ dissociates completely being as strong as acid the moles of hydrogen ions forms are $2 \times 0.05$ too.
Since he initial concentration of the acid is in 100 $ml$ and its diluted to 1 litre, there is a decrease in the concentration by 10 times
Thus the new concentration being $\dfrac{{0.05}}{{10}} \Rightarrow 0.005$
So calculating $pH$

$pH = - \log [{H^ + }]$

$\Rightarrow pH = - \log (2 \times 0.005)$

$\Rightarrow pH = 2$

So, the correct answer is Option A.

Note: The $pH$ of a sample can be tested physically through many ways like the Litmus paper and the $pH$ paper. The $pH$ paper gives different colours based on the different $pH$ of different samples while the litmus paper only allows us to know if the sample is either basic or acidic. It changes the colour to blue if the sample is basic while it changes to red in case of an acidic sample.