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Brut [27]
2 years ago
8

Wasim and his friends want to test if the solution of a substance is acidic or not. These are the tests that each of them perfor

med. Wasim: Dipped the blue litmus in the solution Akram: Dipped the red litmus in the solution Neha: Dipped both the blue and the red litmus in the solution one after other. Who among them performed a proper test ?
Chemistry
1 answer:
alex41 [277]2 years ago
3 0

Answer:

Wasim and Nehra did the Proper testing  

Explanation:

In acidic condition Blue litmus turns red

In basic condition Red litmus turns Blue

In order to test an Acidic solution we need Blue litmus.

Wasim Dipped the blue litmus in the solution : it will turn red and shows acid

Akram  Dipped the red litmus in the solution : No it will not show, it's a test for base

Neha: Dipped both the blue and the red litmus in the solution one after other.

          Blue litmus will turn red which will show Acidic nature, after that Adding Red litmus in it will not show any change as Red litmus only changes in Basic condition.

Therefore,  Wasim and Nehra did the Proper  testing  

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This rule is used to determine number of electrons in particular shell.

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How many grams of sulfuric acid is needed to neutralize 380 ml of solution with pH = 8.94
erma4kov [3.2K]

Answer : The mass of sulfuric acid needed is 16.23\times 10^{-5}g.

Solution : Given,

pH = 8.94

Volume of solution = 380 ml = 380\times 10^{-3}      (1ml=10^{-3}L)

Molar mass of sulfuric acid = 98.079 g/mole

As we know,

pH+pOH=14\\pOH=14-8.94=5.06

pOH=-log[OH^-]

5.06=-log[OH^-]

[OH^-]=0.00000871=8.71\times 10^{-6}mole/L

Now we have to calculate the moles of OH^-.

Formula used : Moles=Concentration\times Volume

\text{ Moles of }[OH^-]=\text{ Concentration of }[OH^-]\times Volume\\\text{ Moles of }[OH^-]=(8.71\times 10^{-6}mole/L)\times (380\times 10^{-3}L)=3309.8\times 10^{-9}moles

For neutralization, equal number of moles of H^+ ions will neutralize same number of OH^- ions.

\text{ Moles of }[OH^-]=\text{ Moles of }[H^+]=3309.8\times 10^{-9}moles

As, H_2SO_4\rightarrow 2H^++SO^{2-}_4

From this reaction, we conclude that

2 moles of H^+ ion is given by the 1 mole of H_2SO_4

3309.8\times 10^{-9} moles of H^+ ion is given by \frac{3309.8\times 10^{-9}}{2}=1654.9\times 10^{-9} moles of H_2SO_4

Now we have to calculate the mass of sulfuric acid.

Mass of sulfuric acid = Moles of H_2SO_4 × Molar mass of sulfuric acid

Mass of sulfuric acid = (1654.9\times 10^{-9}moles)\times (98.079g/mole)=162310.94\times 10^{-9}=16.23\times 10^{-5}g

Therefore, the mass of sulfuric acid needed is 16.23\times 10^{-5}g.

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3 years ago
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