What is #[HO^-]# of a solution that is #6.80xx10^-2*mol*L^-1# with respect to #HBr(aq)#?

Question

Jeremiah Adams · Accepted Answer

#[HO^-]=1.47xx10^-13*mol*L^-1#

We examine the water's hydrolysis reaction in standard circumstances:

#2H_2O(l) rightleftharpoons H_3O^+ + HO^-#

With very precise knowledge of the extent of this equilibrium, we find (under standard conditions) that:

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

The given equation is an #"equation"# (duhh!); we may divide it, multiply it, add to it, PROVIDED that we do it consistently to both left hand and right sides of the equation. One thing we can do is to take #"logarithms to the base 10"# of both sides, and follow the standard rules of arithmetic:

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

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

After reorganization:

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

Now using the definition of a logarithm, when I write #log_ab=c#, this means that #a^c=b#. Perhaps you should review the logarithmic function in your mathematics text.

So the left hand side of the equation, #=-log_10(10^-14)=-14#, i.e.

#-log_10(10^-14)=14#

Thus, I can rewrite the provided equation to produce:

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

#=-14=-log_10[H_3O^+]- log_10[HO^-]#

But by definition, #-log_10[H_3O^+]=pH#, and

#-log_10[HO^-]=pOH#.

And so finally, we can write under standard conditions that #pH+pOH=14#.

After completing all of this, we can (Finally!) discuss the provided question, which I almost forgot to ask.

We have #[H_3O^+]=6.80xx10^-2*mol*L^-1#, why? Well because the #HBr# acid dissociates completely in aqueous solution according to the following equation:

#HBr(g) + H_2O(l) rarr H_3O^+ + Br^-#

And so #pH=-log_10(6.80xx10^-2)=1.17#,

and #pOH=12.83#.

And so #[HO^-]=10^(-12.83)=1.471xx10^-13*mol*L^-1#.

I realize this was a lot of work for such a small problem; all we've done is apply the proper usage of the logarithmic function here.

The key equation is #pH+pOH=14#.

Naomi Aronson · Answer

undefined To find the hydroxide ion concentration ([OH^-]) of a solution of HBr(aq), you first need to consider the dissociation of HBr in water:

HBr(aq) → H^+(aq) + Br^-(aq)

Since HBr is a strong acid, it fully dissociates in water. Therefore, the concentration of H^+ ions ([H^+]) is equal to the concentration of HBr, which is 6.80x10^-2 M.

Given that water autoionizes to produce equal concentrations of H^+ and OH^- ions, the concentration of OH^- ions is also 6.80x10^-2 M.

Thus, [OH^-] = 6.80x10^-2 M.