Our teacher asked, a few weeks ago, for us to find the pH of 7.65 E-11 M HClO3. We were told that we couldn't find the pH simply by -log[7.65E-11]. He said that we had to account for the water somehow [I think we had to do something with the number 1.0 E-7]. Does anyone know what exactly I have to do with the water [I don't have it written down, but I have that the pH= 7.00] and why/when do I have to do this?
Since the H+ contribution is greater than that of the acid, yes, you have to add them:
[H+]= .000 000 1 + .000 000 000 0765
add them, then do the log. The reason that the log is close to seven is that the water domintes as the contribution to the hydrogen ion.
To find the pH of 7.65E-11 M HClO3, you need to account for the contribution of water to the hydrogen ion concentration. Typically, when the concentration of water is much larger than the concentration of the acid or base, it dominates the contribution to the hydrogen ion. In this case, the concentration of water is 1.0E-7 M.
The formula for pH is -log[H+], where [H+] is the concentration of hydrogen ions. However, in this specific case, you cannot simply use -log[7.65E-11] because you need to consider the contribution of water as well.
To find the pH, you need to add the concentration of hydrogen ions contributed by both the acid (HClO3) and the water. The concentration of hydrogen ions contributed by the acid is 7.65E-11 M, and the concentration of hydrogen ions contributed by water is 1.0E-7 M.
Therefore, you need to add these two concentrations together:
[H+] = 7.65E-11 + 1.0E-7
After adding these concentrations, you can take the negative logarithm to find the pH:
pH = -log([H+])
Now, let's calculate the pH using the given values:
[H+] = 7.65E-11 + 1.0E-7
= 1.00765E-7
pH = -log(1.00765E-7)
≈ 6.997
So, the pH of 7.65E-11 M HClO3 is approximately 6.997, which is close to 7.00.
This approach accounts for the contribution of water and provides a more accurate result for the pH calculation.