Calculate the concentration of `overset(Θ)OH` in the solution of base with `pH`
a. `10.4771` b. `12.301` c. `11.8451`

Explanation

Since the solution are basic, so first calculate `pOH` of the solution and then calculate `[overset(Θ)OH]` accordingly.
a. `pH = 10.4771`
`pOH = 14 - 10.4771 3.5229`
`pOH =- log [overset(Θ)OH] = 3.5229`
`:. log [overset(Θ)OH] =- 3.5229 =- 3 - 0.5229 +1 - 1`
`= bar(4).4771`
`:. [overset(Θ)OH] = "Antilog" (bar(4).4771)`
`= 3 xx 10^(-4) = 0.3 xx 10^(-3) N`
c. `pH = 12.301`
`pOH = 14 - 12.301 = 1.699`
`pOH =- log [overset(Θ)OH] = 1.699`
`:. log [overset(Θ)OH] =- 1.699 =- 1 - 0.699 +1 - 1`
`= bar(2).301`
`:. [overset(Θ)OH] = "Antilog" (bar(2).301)`
`= 2 xx 10^(-2)N = 0.2 xx 10^(-1)N`
d. `pH = 11.8451`
`pOH = 14 - 11.8461 = 2.1549`
`pOH =- log [overset(Θ)OH] =- 2.1549 =- 2 - 0.1549 +1 - 1`
`= bar(3).8451`
`:. [overset(Θ)OH] = "Antilog" (bar(3).8451)`
`= 7 xx 10^(-3) N = 0.7 xx 10^(-2) N`