# 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.

Jul 1, 2017

We know that in aqueous solution.........

#### Explanation:

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

${H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s {H}^{+} + H {O}^{-}$

${H}^{+}$ and $H {O}^{-}$ are the characteristic cation, and characteristic anion of the water solvent. Sometimes, we write ${H}^{+}$ as ${H}_{3} {O}^{+}$, i.e. the $\text{hydronium ion}$. As far as anyone knows this is a cluster of 3-4 (or more) water molecules, with an extra PROTON, ${H}^{+}$ to give ${H}_{7} {O}_{3}^{+}$, or ${H}_{9} {O}_{4}^{+}$, we use ${H}^{+}$ and ${H}_{3} {O}^{+}$ as convenient shorthands.

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

$K = \frac{\left[{H}^{+}\right] \left[H {O}^{-}\right]}{\left[{H}_{2} O \left(l\right)\right]}$

Because $\left[{H}_{2} O\right]$ is so large, it can be removed from the equation, and we get

${K}_{w} = \left[{H}^{+}\right] \left[H {O}^{-}\right] = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$ at $298 \cdot K$. Now back in the day before the advent of electronic calculators, scientists used log tables to the $\text{base e}$ or $\text{base 10}$ to simplify products and quotients of very large and very small numbers.

We DEFINE $p H = - {\log}_{10} \left[{H}_{3} {O}^{+}\right]$, and $p O H = - {\log}_{10} \left[H {O}^{-}\right]$, and thus if we take ${\log}_{10}$ of

${K}_{w} = \left[{H}^{+}\right] \left[H {O}^{-}\right] = \left[{H}_{3} {O}^{+}\right] \left[H {O}^{-}\right] = {10}^{- 14}$.....

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

${\log}_{10} {K}_{w} = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right] = {\log}_{10} \left({10}^{-} 14\right)$

And on rearrangement,

$- 14 = {\log}_{10} \left[{H}_{3} {O}^{+}\right] + {\log}_{10} \left[H {O}^{-}\right]$, and multiplying thru by $- 1$..

$14 = {\underbrace{- {\log}_{10} \left[{H}_{3} {O}^{+}\right]}}_{p H} {\underbrace{- {\log}_{10} \left[H {O}^{-}\right]}}_{p O H}$

And finally our defining relationship......$p H + p O H = 14$. 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.

$\text{Solution A}$, $\left[{H}^{+}\right] = 3.7 \times {10}^{-} 8 \cdot m o l \cdot {L}^{-} 1$; $p H = - {\log}_{10} \left(3.7 \times {10}^{-} 8\right) = 7.43$; $p O H = 14 - 7.43 = 6.57$. And $\left[H {O}^{-}\right] = {10}^{- 6.57} = 2.70 \times {10}^{-} 7 \cdot m o l \cdot {L}^{-} 1$.

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

One final caveat: I stress that ${H}^{+}$ is equivalent to ${H}_{3} {O}^{+}$; which expression you use is a matter of personal preference, but be consistent.

Jul 1, 2017

Here's what I get.

#### Explanation:

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

$\text{pH = 14.00 - pH = 14.00 - 6.57 = 7.43}$

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

$\text{pOH = 14.00 - 10.65 = 3.35}$

Solution C

$\text{pH = 8.67}$

$\text{pOH = 14.00 - 8.67 = 5.33}$

["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

$\text{pOH = 12.72}$

$\text{pH = 14.00 - 12.72 = 1.28}$

["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"