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.#

2 Answers
Jul 1, 2017

Answer:

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

Explanation:

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.

Jul 1, 2017

Answer:

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"#