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To find the pOH of solution of $\pu{0.42 M}$ of $\ce{NaHCO3}$, given a $\ce{Kb}$ of $\pu{2.4E-8}$, my solution manual does the following steps:

  1. Reaction is similar to : $\ce{HCO3- + H2O <=> H2CO3 + OH-}$
  2. $\displaystyle \ce{Kb} = \pu{2.4E-8} = \ce{\frac{[OH-][H2CO3]}{[HCO3-]}}$
  3. $\displaystyle \ce{Kb} = \pu{2.4E-8} = \ce{\frac{[OH]^2}{[0.42]}}$ , as we have $\pu{0.42 M}$ of $\ce{NaHCO3}$ so approximately $\pu{0.42 M}$ $\ce{HCO3-}$, and $\ce{[OH-] = [H2CO3]}$
  4. Solve equation for $\ce{[OH-]}$, giving $\ce{[OH-]} = \pu{10^{-4}}$ and hence $\ce{pOH} = 4$

What I don't understand is: Given the equation $$\ce{HCO3- + H2O <=> H2CO3 + OH-},$$ doesn't this show $\ce{HCO3} = \ce{H2O} = \ce{H2CO3} =\ce{OH-}$, as they are all in a $\mathrm{1:1:1:1}$ molar ratio in the equation.

So wouldn't $\ce{OH-}$ also just equal $\pu{0.42 M}$?

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    $\begingroup$ No, it doesn't show that, much like it doesn't show that you have precisely one molecule (or one mole, or one anything) of H2O in your reaction based on the fact that there is one H2O in the equation. $\endgroup$ – Ivan Neretin Apr 12 '17 at 10:32
  • $\begingroup$ Thank you. I must have really misunderstood something when learning this topic. I will have to do some more reading to make understand why your comment is true. $\endgroup$ – K-Feldspar Apr 12 '17 at 10:37
  • $\begingroup$ Could you please point out why we can take [HCO3-]=0.42, but not [OH-]=0.42? We are only told that we have 0.42 M of NaHCO3. $\endgroup$ – K-Feldspar Apr 12 '17 at 10:40
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    $\begingroup$ We can't. We only know that there were 0.42 M of NaHCO3 before the reaction. Then again, judging from experience and looking at the constant, we may say that the reaction occurs to a very little extent, so in the end not much on this NaHCO3 is spent, and its concentration is still almost 0.42M. $\endgroup$ – Ivan Neretin Apr 12 '17 at 11:04
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    $\begingroup$ Change "dissociates" to "hydrolyzes". Other than that, you are right. $\endgroup$ – Ivan Neretin Apr 12 '17 at 20:15
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Your misunderstanding appears to be some confusion between stoichiometry and equilibrium concentrations. The chemical equation describes the stoichiometry or the ratio of the species in the reaction. In other words, each time the reaction happens, the equation describes how many of each reactant are consumed and how many are produced. In your example:

$$\ce{HCO3- + H2O <=> H2CO3 + OH-}$$

Each time this reaction occurs in the forward direction, one bicarbonate anion and one water molecule react to form one carbonic acid molecule and one hydroxide anion. In the reverse direction, one carbonic acid molecule and one hydroxide anion react to form one bicarbonate anion and one water molecule.

Since this reaction is reversible, the equilibrium constant tells you something about the extent of the reaction. In other words, what portion of the overall system is sitting on the reactant side and what portion sis sitting on the product side. Put another way, the system is at equilibrium when the rates of the forward and reverse reactions are equal. These rates are dependent on the concentrations of the species on the reactant and product sides of the equation and on rate constants.

$$\mathrm{rate_{forward}=rate_{reverse}}\\ k_\mathrm{f}[\ce{HCO3-}][\ce{H2O}]=k_\mathrm{r}[\ce{H2CO3}][\ce{OH-}]$$

The equilibrium constant is the ratio of the rate constants:

$$K=\frac{k_\mathrm{r}}{k_\mathrm{f}}=\frac{[\ce{H2CO3}][\ce{OH-}]}{[\ce{HCO3-}][\ce{H2O}]}$$

The equilibrium constant is thus a law of mass action expression for the situation where the forward and reverse rates are equal. At this state, both forward and reverse reactions are occurring following the stoichiometric ratios, but the concentrations of each species has reached a steady state.

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  • $\begingroup$ Thanks Ben. I think I understand now: Recapping what you and other answer said: The reaction has a low Kb, so little of it dissociates, so HCO3- approx. = NaHCO3 = 0.42M (approx. equal to our starting point as Kb is so low - and so products <<<<< reactants). [OH-] = [H2CO3], as they are both produced 1:1 as both have a coefficient of 1 and both appear as products. BUT we can't say [HCO3-]=[OH-] = [H2CO3] because Kb is not =1, and so 1 mole of HCO3- does not get fully used up and (in combination with H2O) become 1 mole of H2CO3+ 1 mole of [OH-]. Does that sound right? Thanks $\endgroup$ – K-Feldspar Apr 12 '17 at 20:13
  • $\begingroup$ Hi Ben, I have a follow up to this. For the reaction NH3 + H2O <=> NH4+ + OH- Given 0.5 mole of amoonia dissolved in 250 mL of water, I am told [OH] is 2 mol/L Why is this the case i.e. 2 mol/L [NH3] is taken to give 2 mol/L [OH-] but in my original question we were not allowed to do this, because the reactants don't necessarily all get used up to form products. $\endgroup$ – K-Feldspar Jan 11 '18 at 1:40
  • $\begingroup$ Note I have opened another Q which asks about this chemistry.stackexchange.com/questions/88663/… $\endgroup$ – K-Feldspar Jan 11 '18 at 2:46

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