What are `pH`, and `pOH` for solutions where...:? `A.` `[H^+]=3.7xx10^-8*mol*L^-1;` `B.` `[HO^-]=4.5xx10^-4*mol*L^-1;` `C.` Where `pH=8.67`, what are `[H_3O^+]` and `[HO^-];` `D.` `pOH=12.72.`

We know that is aqueous solution the following equilibrium operates......

`H_2O(l) rightleftharpoonsH^(+) + HO^-`

`H^+` and `HO^-` are the characteristic cation, and characteristic anion of the water solvent. Sometimes, we write `H^+` as `H_3O^+`, i.e. the `"hydronium ion"`. As far as anyone knows this is a cluster of 3-4 (or more) water molecules, with an extra PROTON, `H^+` to give `H_7O_3^+`, or `H_9O_4^+`, we use `H^+` and `H_3O^+` as convenient shorthands.

As with any equilibrium we can quantify it......

`K=([H^+][HO^-])/[[H_2O(l)]]`

Because `[H_2O]` is so large, it can be removed from the equation, and we get

`K_w=[H^+][HO^-]=[H_3O^+][HO^-]=10^(-14)` at `298*K`. Now back in the day before the advent of electronic calculators, scientists used log tables to the `"base e"` or `"base 10"` to simplify products and quotients of very large and very small numbers.

We DEFINE `pH=-log_10[H_3O^+]`, and `pOH=-log_10[HO^-]`, and thus if we take `log_10` of

`K_w=[H^+][HO^-]=[H_3O^+][HO^-]=10^(-14)`.....

..............we gets.................

`log_10K_w=log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)`

And on rearrangement,

`-14=log_10[H_3O^+]+log_10[HO^-]`, and multiplying thru by `-1`..

`14=underbrace(-log_10[H_3O^+])_(pH)underbrace(-log_10[HO^-])_(pOH)`

And finally our defining relationship......`pH+pOH=14`. Wheww!

And so we can at last address your question. I will do the first row in the table. I suggest that you have a go at the rest yourself.

`"Solution A"`, `[H^+]=3.7xx10^-8*mol*L^-1`; `pH=-log_10(3.7xx10^-8)=7.43`; `pOH=14-7.43=6.57`. And `[HO^-]=10^(-6.57)=2.70xx10^-7*mol*L^-1`.

Over to you for `B` and `C` and `D` hoss............

One final caveat: I stress that `H^+` is equivalent to `H_3O^+`; which expression you use is a matter of personal preference, but be consistent.

Solution A

`["H"^"+"] = 3.7 × 10^"-8"color(white)(l) "mol/L"`

`["OH"^"-"] = K_text(w) /(["H"^"+"]) = (1.00 × 10^"-14")/(3.7 × 10^"-8") "mol/L" = 2.7 × 10^"-7"color(white)(l) "mol/L"`

`"pOH" = -log["OH"^"-"] = 6.57`

`"pH = 14.00 - pH = 14.00 - 6.57 = 7.43"`

Solution B

`["OH"^"-"] = "0.000 45 mol/L"`

`["H"^"+"] = (1.00 × 10^"-14")/"0.000 45" "mol/L" = 2.2 × 10^"-11"color(white)(l) "mol/L"`

`"pH" = -log(2.2 × 10^"-11") = 10.65`

`"pOH = 14.00 - 10.65 = 3.35"`

Solution C

`"pH = 8.67"`

`"pOH = 14.00 - 8.67 = 5.33"`

`["OH"^"-"] = 10^"-5.33"color(white)(l)"mol/L" = 4.7 × 10^"-6" color(white)(l)"mol/L"`

`["H"^"+"] = 10^"-8.67"color(white)(l)"mol/L" = 2.14 × 10^"-9" color(white)(l)"mol/L"`

Solution D

`"pOH = 12.72"`

`"pH = 14.00 - 12.72 = 1.28"`

`["H"^"+"] = 10^"-1.28"color(white)(l)"mol/L" = "0.052 mol/L"`

`["OH"^"-"] = 10^"-12.72"color(white)(l)"mol/L" = 1.91 × 10^"-13" color(white)(l)"mol/L"`