Calculate the pH at 298K of solutions having the following ion concentrations?

A. [H+]=1.0 x 10^-4M
B. [H+]=5.8 X 10^-11M

for A do you subtract 1.0 X 10^-14 from H+]=1.0 x 10^-4M

For A, no. See an example of both here.

http://www.jiskha.com/display.cgi?id=1268100336

To calculate the pH of a solution, you can use the formula:

pH = -log[H+]

Where [H+] refers to the concentration of hydrogen ions in the solution.

For the first case (A), the given concentration is [H+] = 1.0 x 10^-4 M. To calculate the pH, you need to take the negative logarithm (base 10) of this concentration:

pH = -log(1.0 x 10^-4) = -(-4) = 4

Therefore, the pH of solution A is 4.

Now, in regards to your second question, you mentioned subtracting 1.0 x 10^-14. It seems you are referring to the value of the ion product of water, which is Kw = [H+][OH-] = 1.0 x 10^-14 at 298K. However, this value is not directly involved in calculating the pH of a solution. The ion product of water is used to relate the concentrations of hydrogen ions ([H+]) and hydroxide ions ([OH-]) in a neutral solution.

To calculate the pH of a solution, you only need the concentration of hydrogen ions ([H+]). Simply take the negative logarithm of this concentration using the formula mentioned earlier.

To calculate the pH of a solution, you can use the equation: pH = -log[H+].

For solution A, where [H+] = 1.0 x 10^-4 M, you can directly plug this value into the equation:
pH = -log(1.0 x 10^-4)

Using a calculator, this would simplify to:
pH = -log(0.0001)
pH = 4

So, the pH for solution A is 4.

Now, for solution B, where [H+] = 5.8 x 10^-11 M, you can follow the same process:
pH = -log(5.8 x 10^-11)

Again, using a calculator, this would simplify to:
pH = -log(0.000000000058)
pH ≈ 10.24

So, the pH for solution B is approximately 10.24.