Acids / Bases question

[fingerscrossedd]

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From EK "In class exams" Chapter 6 Chem:

133. What is the pH of a 5.0 X 10^-8 M aqueous solution of H2SO4 at room temperature?

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EK explains this with:

"Water is the main contributor of H+. Add the 5E-8 ions contributed by H2SO4 to the 1E-7 ions contributed by water. This leaves 1.5E-7 H+ ions."



Okay so...since H2SO4 is a strong acid, concentration of H+ = initial concentration of acid, right? That's why this approach works? It just sort of threw me for a loop when I first read it, and I'm only now beginning to understand it.

Thanks! 😍😍😍

 

[chem4ever]

From EK "In class exams" Chapter 6 Chem:

133. What is the pH of a 5.0 X 10^-8 M aqueous solution of H2SO4 at room temperature?

--------

EK explains this with:

"Water is the main contributor of H+. Add the 5E-8 ions contributed by H2SO4 to the 1E-7 ions contributed by water. This leaves 1.5E-7 H+ ions."



Okay so...since H2SO4 is a strong acid, concentration of H+ = initial concentration of acid, right? That's why this approach works? It just sort of threw me for a loop when I first read it, and I'm only now beginning to understand it.

Thanks! 😍😍😍

Usually, when dealing with strong acids, you don't need to take into account the acidity of water because the H+ from the strong acid is much greater than the H+ from the water so the H+ from the water is negligible. For example, a 1 x 10^-2 M H2SO4 will have a pH of 2 even though there is 1 x 10^-7 M H+ from water. Technically, you need to add both sources of H+ so you would get 1 x 10^-2 M (from the H2SO4) + 1 x 10^-7 M (from the water) = 1.00001 x 10^-2 M H+. As you can see the 0.00001 is not contributing much so you can ignore it and you get a pH of 2.

However, when the molarity of the strong acid is less (or very close) to the H+ from water (1 x 10^-7), then it is no longer negligible. In this case, the molarity of the H2SO4 was less that the molarity of H+ from water. So you need to add them to get the total molarity of the H+ and then calculate the pH.

So in this case, we get 5 x 10^-8 M (from H2SO4) + 1 x 10^-7 (from water) = 1.5 x 10^-7.

pH = -log (1.5 x 10^-7) = 6.82

 

[fingerscrossedd]

Usually, when dealing with strong acids, you don't need to take into account the acidity of water because the H+ from the strong acid is much greater than the H+ from the water so the H+ from the water is negligible. For example, a 1 x 10^-2 M H2SO4 will have a pH of 2 even though there is 1 x 10^-7 M H+ from water. Technically, you need to add both sources of H+ so you would get 1 x 10^-2 M (from the H2SO4) + 1 x 10^-7 M (from the water) = 1.00001 x 10^-2 M H+. As you can see the 0.00001 is not contributing much so you can ignore it and you get a pH of 2.

However, when the molarity of the strong acid is less (or very close) to the H+ from water (1 x 10^-7), then it is no longer negligible. In this case, the molarity of the H2SO4 was less that the molarity of H+ from water. So you need to add them to get the total molarity of the H+ and then calculate the pH.

So in this case, we get 5 x 10^-8 M (from H2SO4) + 1 x 10^-7 (from water) = 1.5 x 10^-7.

pH = -log (1.5 x 10^-7) = 6.82

Wow, super helpful- thanks a lot!!!😍

 

[adrakdavra]

I love chem nerds