Which of the following solution will have `pH = 13`?

Options

A. `2g NaOH "in" 500 mL` solution.
B. `100 mL` solution fo `0.05M Ca(OH)_(2)`.
C. `100mL` solution of `1.0N Ca(OH)_(2)`.
D. `4gNaOH` in `500 mL` solution.

Answer

A, B, C

Explanation

a. `M = (W_(2)xx1000)/(Mw_(2)xxV_(sol)("in mL")) = (2xx1000)/(40xx500) = 0.1M`
`:. [NaOH] = [overset(Theta)OH] = 0.1 = 10^(-1)M`
`pOH =1, pH = 13`
b. `0.05 M Ca(OH)_(2) =2xx 0.05N Ca(OH)_(2)`
`:. [overset(Theta)OH] = 2xx 0.05 = 0.1 = 10^(-1)M`
`{:(pOH=1,pH=13):}`
c. `100 "mL of" 0.1 NaC (OH)_(2)`,
`:. [overset(Theta)OH] = 0.1N = 10^(-1)N`
`{:(pOH=1,pH=13):}`
d. `M = (4xx1000)/(40xx500) = (1)/(5) = 0.2M`
`[overset(Theta)OH] = 0.2M`
`pOH =- log[0.2] =- log 2+ log 10 =- 0.3010 +1 = 0.699`
`pH = 14 - 0.699 = 13.301`
`pH = 14 - 0.699 = 13.301`
So the answer (a), (b) and (c) are correct.