I am in the process of calculating the hydrogen ion concentration in a .25litre solution of sulfuric acid,.21g of sulfuric acid was disolved in water to make the solution. I have calculated the molarity of the acid to be 2.14x10^-3 mol per litre.

Question

I am in the process of calculating the hydrogen ion concentration in a .25litre solution of sulfuric acid,.21g of sulfuric acid was disolved in water to make the solution. I have calculated the molarity of the acid to be 2.14x10^-3 mol per litre.

I also need help in calculating the pH of the acid soulution.

Well, if the H ion is twice that concentration..

pH= -log(2*2.14*10^-3)
grab your calculator.

I think you should check your calculations. I THINK you have 0.21/98 = 2.14 x 10^-3 mols but that is in 0.25 L. You want mols/L.

I have calculated the mols/L by multiplying .25ltrs and .21g by 4. This gives .84mols per litre. can this be confirmed for me please.

Answer 1

To confirm your calculation, we can use stoichiometry to calculate the number of moles of sulfuric acid in the solution.

First, we need to determine the molar mass of sulfuric acid. It consists of one sulfur atom (S) with a molar mass of approximately 32.07 g/mol and four oxygen atoms (O) with a molar mass of approximately 16.00 g/mol each. The molar mass of sulfuric acid can be calculated as follows:

Molar mass of sulfuric acid = (1 * molar mass of sulfur) + (4 * molar mass of oxygen)
= (1 * 32.07 g/mol) + (4 * 16.00 g/mol)
= 98.09 g/mol

Now, we can calculate the number of moles of sulfuric acid in the solution using the given mass and molar mass:

Number of moles = mass of sulfuric acid / molar mass
= 0.21 g / 98.09 g/mol
= 0.002143 mol

So, you were correct in calculating the moles of sulfuric acid as 2.143 x 10^-3 mol (rounded to four decimal places).

To calculate the pH of the acid solution, we need to use the equation pH = -log[H+], where [H+] is the concentration of hydrogen ions in moles per liter (M).

Given that the sulfuric acid is a strong acid, it fully dissociates in water, resulting in twice the concentration of hydrogen ions. The concentration of hydrogen ions is therefore 2 * 2.143 x 10^-3 M = 4.286 x 10^-3 M.

Now, we can plug this value into the pH equation:

pH = -log(4.286 x 10^-3)
≈ 2.37

So, the pH of the sulfuric acid solution is approximately 2.37.