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

Your calculations for determining the molarity of the original H2SO4 are correct. The molar mass of H2SO4 is indeed 98 g/mol. Therefore, the moles of H2SO4 in the 0.21 g sample are 2.1 x 10^-3 mol. Since the moles of H+ are twice that of the original acid, the original concentration is double. Therefore, the molarity of the original H2SO4 is 4.2 x 10^-3 M (to 2 significant figures).

To calculate the hydrogen ion concentration (in mol/L), you need to consider the dissociation equation for sulfuric acid:

H2SO4(aq) -> 2 H+(aq) + SO4^2-(aq)

From the equation, you can see that one mole of H2SO4 produces two moles of H+ ions. Therefore, the concentration of H+ ions is twice the molarity of the original H2SO4 solution.

In this case, the molarity of the original H2SO4 is 4.2 x 10^-3 M. So, the hydrogen ion concentration is 2 times that, which is 8.4 x 10^-3 M.

Now, to calculate the pH of the sulfuric acid solution, you can use the formula pH = -log10(concentration of H+).

Plugging in the value for the concentration of H+ (8.4 x 10^-3 M), you get:

pH = -log10(8.4 x 10^-3)

Using a calculator, you can find the logarithm and convert it to the pH value. The pH value should be the nearest whole number.

Therefore, the pH of the sulfuric acid solution is approximately 2 (nearest whole number).