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Question: One lit. of aqueous solution of \[{H_2}S{O_4}\] contains 4.9 \(g\) of the acid. 100 \(ml\) of this s...

One lit. of aqueous solution of H2SO4{H_2}S{O_4} contains 4.9 gg of the acid. 100 mlml of this solution is taken in a 1 lit. flask and diluted with water up to the mark. The pHpH of the dilute solution is
A. 2.0
B. 2.301
C. 1.699
D. 3.699

Explanation

Solution

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.

Complete step by step answer:
pHpH of a sample is the power of the hydrogen ions that the sample gives. The pHpH 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 H2SO4{H_2}S{O_4} which is an acid so it is most likely to have the pHpH below 7.

H2SO4{H_2}S{O_4} dissociates in the solution as

H2SO42H++SO42{H_2}S{O_4}\xrightarrow[{}]{}2{H^ + } + SO_4^{2 - }

In order to find the pHpH we need to calculate the concentration of H2SO4{H_2}S{O_4} in the solution which gives the hydrogen ion.

Moles=massmolar massMoles = \dfrac{\text{mass}}{\text{molar mass}},

Molar mass of H2SO4{H_2}S{O_4} can be calculated by adding the molar mass of its consisting atoms as

2(MassofH)+massofS+4(massofO)2(Mass\,of\,H) + mass\,of\,S + 4(mass\,of\,O)
98\Rightarrow 98

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

moles=4.998moles = \dfrac{{4.9}}{{98}}

moles=0.05 \Rightarrow moles = 0.05

So, since the H2SO4{H_2}S{O_4} dissociates completely being as strong as acid the moles of hydrogen ions forms are 2×0.052 \times 0.05 too.
Since he initial concentration of the acid is in 100 mlml and its diluted to 1 litre, there is a decrease in the concentration by 10 times
Thus the new concentration being 0.05100.005\dfrac{{0.05}}{{10}} \Rightarrow 0.005
So calculating pHpH

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

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

pH=2 \Rightarrow pH = 2

So, the correct answer is Option A.

Note: The pHpH of a sample can be tested physically through many ways like the Litmus paper and the pHpH paper. The pHpH paper gives different colours based on the different pHpH 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.