Water has a pH of 4.9? Calculate H3O and OH

1. H3O= 10^4.9=1.25*10^-5
2. OH= 14- 4.9= 9.1

ignore this. I saw your earlier post thank you.

Thanks for letting me know.

To calculate the concentration of H3O+ (hydronium ions) and OH- (hydroxide ions) in a solution with a pH of 4.9, you can use the formula:

pH = -log[H3O+]

To find the concentration of H3O+:

1. Take the antilog of the pH value: H3O+ = 10^(-pH)

In this case, the pH is 4.9, so the concentration of H3O+ becomes:

H3O+ = 10^(4.9) = 1.25 * 10^(-5) mol/L

To find the concentration of OH-:

1. Use the fact that in pure water, the product of the concentration of H3O+ and OH- ions is always equal to 1 x 10^(-14) at 25°C, as expressed in the equation: [H3O+][OH-] = 1 x 10^(-14)

2. Rearrange the equation to solve for OH-: OH- = 1 x 10^(-14) / H3O+

In this case, we can substitute the concentration of H3O+ we calculated earlier:

OH- = 1 x 10^(-14) / (1.25 * 10^(-5))
= 8 x 10^8 mol/L

Therefore, the concentration of H3O+ is 1.25 * 10^(-5) mol/L, and the concentration of OH- is 8 x 10^8 mol/L.