What are $p H$ $pH$ , and $p O H$ $pOH$ for solutions where...:? $A .$ $A.$ $[H^{+}] = 3.7 \times 10^{- 8} \cdot m o l \cdot L^{- 1};$ $[H^+]=3.7xx10^-8molL^-1;$ $B .$ $B.$ $[H O^{-}] = 4.5 \times 10^{- 4} \cdot m o l \cdot L^{- 1};$ $[HO^-]=4.5xx10^-4molL^-1;$ $C .$ $C.$ Where $p H = 8.67$ $pH=8.67$ , what are $[H_{3} O^{+}]$ $[H_3O^+]$ and $[H O^{-}];$ $[HO^-];$ $D .$ $D.$ $p O H = 12.72 .$ $pOH=12.72.$

Question

What are $p H$ $pH$ , and $p O H$ $pOH$ for solutions where...:? $A .$ $A.$ $[H^{+}] = 3.7 \times 10^{- 8} \cdot m o l \cdot L^{- 1};$ $[H^+]=3.7xx10^-8molL^-1;$ $B .$ $B.$ $[H O^{-}] = 4.5 \times 10^{- 4} \cdot m o l \cdot L^{- 1};$ $[HO^-]=4.5xx10^-4molL^-1;$ $C .$ $C.$ Where $p H = 8.67$ $pH=8.67$ , what are $[H_{3} O^{+}]$ $[H_3O^+]$ and $[H O^{-}];$ $[HO^-];$ $D .$ $D.$ $p O H = 12.72 .$ $pOH=12.72.$

2 Answers

Impact of this question

7300 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-01T08:24:28

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

Explanation:

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

$H_{2} O (l) ⇌ H^{+} + H O^{-}$ $H_2O(l) rightleftharpoonsH^(+) + HO^-$

$H^{+}$ $H^+$ and $H O^{-}$ $HO^-$ are the characteristic cation, and characteristic anion of the water solvent. Sometimes, we write $H^{+}$ $H^+$ as $H_{3} O^{+}$ $H_3O^+$ , i.e. the $hydronium ion$ $"hydronium ion"$ . As far as anyone knows this is a cluster of 3-4 (or more) water molecules, with an extra PROTON, $H^{+}$ $H^+$ to give $H_{7} O_{3}^{+}$ $H_7O_3^+$ , or $H_{9} O_{4}^{+}$ $H_9O_4^+$ , we use $H^{+}$ $H^+$ and $H_{3} O^{+}$ $H_3O^+$ as convenient shorthands.

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

$K = \frac{[H^{+}] [H O^{-}]}{[H_{2} O (l)]}$ $K=([H^+][HO^-])/[[H_2O(l)]]$

Because $[H_{2} O]$ $[H_2O]$ is so large, it can be removed from the equation, and we get

$K_{w} = [H^{+}] [H O^{-}] = [H_{3} O^{+}] [H O^{-}] = 10^{- 14}$ $K_w=[H^+][HO^-]=[H_3O^+][HO^-]=10^(-14)$ at $298 \cdot K$ $298*K$ . Now back in the day before the advent of electronic calculators, scientists used log tables to the $base e$ $"base e"$ or $base 10$ $"base 10"$ to simplify products and quotients of very large and very small numbers.

We DEFINE $p H = - {log}_{10} [H_{3} O^{+}]$ $pH=-log_10[H_3O^+]$ , and $p O H = - {log}_{10} [H O^{-}]$ $pOH=-log_10[HO^-]$ , and thus if we take ${log}_{10}$ $log_10$ of

$K_{w} = [H^{+}] [H O^{-}] = [H_{3} O^{+}] [H O^{-}] = 10^{- 14}$ $K_w=[H^+][HO^-]=[H_3O^+][HO^-]=10^(-14)$ .....

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

${log}_{10} K_{w} = {log}_{10} [H_{3} O^{+}] + {log}_{10} [H O^{-}] = {log}_{10} (10^{- 14})$ $log_10K_w=log_10[H_3O^+]+log_10[HO^-]=log_10(10^-14)$

And on rearrangement,

$- 14 = {log}_{10} [H_{3} O^{+}] + {log}_{10} [H O^{-}]$ $-14=log_10[H_3O^+]+log_10[HO^-]$ , and multiplying thru by $- 1$ $-1$ ..

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

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.

Answer 2 · 2017-07-01T16:25:23

Here's what I get.

Explanation:

Solution A

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

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

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

$"pH = 14.00 - pH = 14.00 - 6.57 = 7.43"$

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

$"pOH = 12.72"$

$"pH = 14.00 - 12.72 = 1.28"$

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

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

Science

Math

Humanities

... and beyond

2 Answers

Explanation:

Explanation:

Related questions

Impact of this question