Given the concentration of either hydrogen ion or hydroxide ion, use the ion product constant of water to calculate the concentration of the other ion at 298K
A. [H+]= 1.0*10^-4
B. [OH-]=1.3*10^-2
for A do you subtract 1.0*10^-14?
and for B do you divide 1.0*10^-14 and 1.3*10^-2
To calculate the concentration of the other ion (hydroxide or hydrogen) using the ion product constant of water at 298K, you need to apply the principle of the ion product constant (Kw) and use the given concentration.
The ion product constant of water (Kw) is a constant value at a given temperature that relates the concentrations of hydrogen ions ([H+]) and hydroxide ions ([OH-]) in water. At 298K, the value of Kw is approximately 1.0 x 10^(-14) mol^2/L^2.
Let's go through the calculations for both options:
A. [H+] = 1.0 x 10^(-4) mol/L:
For this case, you want to calculate the concentration of hydroxide ions ([OH-]). Since Kw = [H+][OH-], you can rearrange the equation to solve for [OH-]. Divide both sides of the equation by [H+] to isolate [OH-]:
[OH-] = Kw / [H+]
[OH-] = (1.0 x 10^(-14) mol^2/L^2) / (1.0 x 10^(-4) mol/L)
[OH-] = (1.0 x 10^(-14) mol) / (1.0 x 10^(-4) L)
[OH-] = 1.0 x 10^(-10) mol/L
So, for option A, the concentration of [OH-] is 1.0 x 10^(-10) mol/L.
B. [OH-] = 1.3 x 10^(-2) mol/L:
For this case, you want to calculate the concentration of hydrogen ions ([H+]). Again, since Kw = [H+][OH-], you can rearrange the equation to solve for [H+]. Divide both sides of the equation by [OH-] to isolate [H+]:
[H+] = Kw / [OH-]
[H+] = (1.0 x 10^(-14) mol^2/L^2) / (1.3 x 10^(-2) mol/L)
[H+] = (1.0 x 10^(-14) mol) / (1.3 x 10^(-2) L)
[H+] ≈ 7.7 x 10^(-13) mol/L
So, for option B, the concentration of [H+] is approximately 7.7 x 10^(-13) mol/L.
To summarize:
- For option A, you divide the ion product constant (Kw) by the hydrogen ion concentration ([H+]) to get the hydroxide ion concentration ([OH-]).
- For option B, you divide the ion product constant (Kw) by the hydroxide ion concentration ([OH-]) to get the hydrogen ion concentration ([H+]).
For A, no. (OH^-) = 1 x 10^-14/1 x 10^-4.
For B, (H^+) = 1 x 10^-14/1.3 x 10^-2.