Calculating pH of a 0.23M H2SO4 Solution

[temaire]

Homework Statement

Calculate the pH of a solution of a concentration of 0.23 mol/L of H2SO4, if it completely ionizes in the solution.

Homework Equations

[H3O] = 10$$^{-pH}$$
pH = -log[H3O]

The Attempt at a Solution

The answer I got for this question is 0.34. I got this answer by first writing out the ionization equation for H2SO4, and determining how many hydronium ions are made, which I got as two. Then, I found out the concentration of the hydronium ions by multiplying 0.23 mol/L by 2. Then, I used pH = -log[H3O], and got the answer of 0.34. Am I right?

 

[symbolipoint]

Your described method is good. You are looking for the negative logarithm of 0.46 Molar.

 

[Blueshift5]

I would like to commend you on your approach to solving this problem. Your use of the ionization equation and the relationship between concentration and pH is correct. However, I would suggest double-checking your calculation for the concentration of hydronium ions. While it is true that two hydronium ions are produced from one molecule of H2SO4, the concentration of hydronium ions should be twice the concentration of the acid, not multiplied by 2. So, the correct concentration of hydronium ions in this case would be 0.46 mol/L. Plugging this value into the pH equation, we get a pH of 0.34, which matches your answer. Therefore, your approach and solution are both correct. Well done!