Chemistry Help.......

Sulfuric Acid (H2SO4) is a strong acid, which dissociates when dissolved in water according to the following equation:

H20SO4(aq)-> 2H+(aq) + SO4 2- (a)

A 0.21g sample of sulfuric acid is dissolved completely in sufficiebt water to make 0.25litre of the final solution.
Calculate the hydrogen ion concentration (in mol 1-1) in this solution.
Answer to scientific notation to an appropriate number of sig fig.

Show steps and explain your reasoning

b)Also what is the pH of the sulfuric acid solution nearet whole number


What is the molarity of the original H2SO4? calculate moles from the mass, then calculate molarity. Notice the moles of H is twice that of the original acid, so double that original concentration. You should have two sig figures.

pH= -log10 (concentration of H)

We will be happy to critique your work.

Does this seems ok?

The molar mass of the sulphuric acid is
(1.01 X 2 + 32.1 + 16.0 x 4) = (2.02 + 32.1 + 64.0) = 98 g (to 2 significant figures).

0.21 g of sulphuric acid contain 0.21/98.1= 2.1x10-3 mol (to 2 significant figures).

Because the moles of H are twice that of the original acid then the original concentration is double. 2.1x10-3 mol x 2 = 4.2 x 10-3


This amount 4.2 x 10-3 is contained in one litre of solution, for 0.25 litres of final solution, the hydrogen concentration is: 4.2 x 10-3 /0.25 l =1.7 x 10-2.

Still not sure about the pH

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asked by Adam

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