Determine the concentration of hydroxide ion in a 4500 ml solution?

Question

Determine the concentration of hydroxide ion in a 4500 ml solution containing 3.78 g. hydrogen bromide

Kw=100 X 10-14 M^2

A. [OH-] = 9.62 X 10^-13 M

B. [OH-]= 2.14 X 10 ^-3 M

C. [OH-]= 4.67 X 10 ^-2 M

D. [OH-]= .0104 M

Answer 1

Beanthai was almost right but she made a noticeable mistake. Hydrogen bromide or hydrobromous acid is a strong acid so it ionizes completely.  First lets find the concentration of the HBr solution:

3.78g of HBr *(mole/80.91g) *(mole/4.5L)= .01038M HBr(aq)

Since HBr is a strong acid, for every mole of HBr ionized there is one mole of H+ produced.

[HBr] = [H+] = .01038M

Kw = [H+]*[OH-]

1*10^-14 = (.01038) * [OH-]

[OH-] = 9.632*10^-13M

Once you look at the question you don't even need to do all this work. Acids have higher concentrations of H+ and lower concentrations of OH-. Bases have higher concentrations of OH- and lower concentrations of H+. All of the answers other than A have high concentrations of OH which means that there bases. This answer can easily be solved by deductive reasoning.

The answer is A.

Answer 2

Ok. in order to find concentration, you use Molarity= mol/L

First find [H+] or hydronium concentration which for strong acids = [HBr] the acid itself.

# of grams divided by Molar Mass

3.78g HBr/80.912g HBr

= .0467 moles (concentration) [H+]

Divide this into Kw because the definition of Kw (for a strong acid) is:

[H+] x [OH-]= Kw

[OH-]= 2.14x10^-13

This makes sense if you take the pH and find it equals 14.

-log(.0467) =1.33 pH

-log(2.14x10^-13)= 12.66 pOH

=14