How is the concentration of H+ lons related to the concentration of OH- ions for a substance?

1 Answer
May 15, 2017

Answer:

Well, we must first specify the water solvent...........and we get under standard conditions, #pH+pOH=14#.

Explanation:

And we know that water undergoes the autoprolysis reaction:

#2H_2OrightleftharpoonsH_3O^+ +HO^-#, we can write,

#K_w=[H_3O^+][HO^-]=10^-14#, under standard conditions. Note that #H_3O^+# and #H^+# can be used fairly interchangeably. The actual species is possibly a cluster of water molecules with AN extra #H^+# associated with it; the proton can tunnel across clusters to give the observed ionic mobility. Likewise, #HO^-#, is a cluster of water molecules with ONE less #H^+#; of course we may isolate actual salts of #HO^-#.

And given #K_w=[H_3O^+][HO^-]=10^-14#, we can take #log_10# of both sides; we remember the definition of the logarithmic function:

#log_(10)0.1=log_(10)10^-1=-1; log_(10)1=log_(10)10^0=0;log_(10)10=+1;log_(10)100=2;.........................................................# and #log_(10)1000=log_(10)10^3=3.#

And so #log_10K_w=log_10([H_3O^+][HO^-])#

#log_10(10^-14)=log_10[H_3O^+]+log_10[HO^-]#,

i.e. #-14=log_10[H_3O^+]+log_10[HO^-]#, multiply each side by #-1#, and thus we get......

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

And thus our defining relationship, the goal of all the previous rigmarole:

#pH+pOH=14#

You will not have to derive this function, but you will use it. Equivalently, in aqueous solution, #pK_a+pK_b=14#.