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.