For each of the following strong base solutions determine OH-, H3O+, pH, and pOH:
1.15×10−2 M Ba(OH)2
2.2×10−4 M KOH
4.9×10−4 M Ca(OH)2
1.15×10−2 M Ba(OH)2
2.2×10−4 M KOH
4.9×10−4 M Ca(OH)2
-
I'm going to do one, since they're quite repetitive. Only thing you have to watch out for is mole ratios.
For example, for Ba(OH)2, it dissociates into Ba+ + 2OH-, meaning the concentration of OH- is twice that of Barium Hydroxide.
With this in mind, the concentration of OH-, represented as [OH-], is .3 or 3.0x10^-1
The equation 10^-14 = [OH-][H3O+] allows us to get [H3O+] since we just found [OH-]
10^-14 = [3x10^-1][H3O+] Solve for [H3O+] by dividing. Concentration of H3O equals 3.33x10^-14.
pH is the negative log of [H3O+] since [H3O+]=[H+] -log(3.33x10^-14) = 13.48
pOH can be found three ways. 10^-14 = pH * pOH, 14 - pH, or you can take the -log of pOH
-log(3.0x10^-1)= .523
FOR Ba(OH)2
[OH-] = 3.0x10^-1
[H3O+] = 3.33x10^-14
pH = 13.48
pOH = .523
For example, for Ba(OH)2, it dissociates into Ba+ + 2OH-, meaning the concentration of OH- is twice that of Barium Hydroxide.
With this in mind, the concentration of OH-, represented as [OH-], is .3 or 3.0x10^-1
The equation 10^-14 = [OH-][H3O+] allows us to get [H3O+] since we just found [OH-]
10^-14 = [3x10^-1][H3O+] Solve for [H3O+] by dividing. Concentration of H3O equals 3.33x10^-14.
pH is the negative log of [H3O+] since [H3O+]=[H+] -log(3.33x10^-14) = 13.48
pOH can be found three ways. 10^-14 = pH * pOH, 14 - pH, or you can take the -log of pOH
-log(3.0x10^-1)= .523
FOR Ba(OH)2
[OH-] = 3.0x10^-1
[H3O+] = 3.33x10^-14
pH = 13.48
pOH = .523