I am performing a simple acid-base titration investigating the efficency of different pure antacid compounds, and in my results, I found out that NaHCO3 neutralized the fastest (using a pH by time graph). Later on, I decided to further prove these results, by calculating pKb values for each base, and out of my three bases (Mg(OH)2, CaCO3, and NaHCO3), NaHCO3 had the highest pKb, which would suggest that it is the weakest base, and wouldn't neutralize the fastest. What should I do in this situation??

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    $\begingroup$ You text does not match the title. What research paper? $\endgroup$ – Karl Mar 11 at 13:10
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    $\begingroup$ Look up the pK values. Edit your question, add a table with the values from literature, and those you calculated from your measurements. Show your calculation. $\endgroup$ – Karl Mar 11 at 13:12
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    $\begingroup$ What do you refer to / how do you qualify «fastest»? This is not a kinetic analysis because (assuming aqueous solution) the neutralization of $\ce{H+}$ by $\ce{OH-}$ to yield $\ce{H2O}$ will be a fast one anyway. Maybe you refer to «to neutralize amount $x$ of acid, amount $y_1$ of antacid (1) is necessary, which is half the amount of $y_2$ using antacid(2)», or a plot of pH value in function of the volume of antacid added. $\endgroup$ – Buttonwood Mar 11 at 13:16
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    $\begingroup$ As Buttonwood says that you can't really think on kinetics. At most you can relate how long there will be a base depending on availability. It can have different grains size (also overcame by the fast reaction, btw) or it can be in a jelly carrier. But this would mean comparing different drugs / formulation not really the bases. $\endgroup$ – Alchimista Mar 11 at 13:23

I think I know what your problem is. Yes, $\ce{NaHCO3}$ is the weakest base of the three, but it is the most soluble in water. Thus, the neutralization can occur throughout the solution. For $\ce{CaCO3}$ and $\ce{Mg(OH)2}$, the rate limiting step is dissolution of the solid. Neutralization only occurs at or near the solid surface.

You also need to be careful mixing kinetic (speed, lower case $k$) and thermodynamic (upper case $K_b$) parameters. The thermodynamic parameters tell you how likely a reaction is, not how fast it will proceed.

However, in this case, we can also see that the reaction with $\ce{NaHCO3}$ is most likely. The acid-base equilibrium is not the only equilibrium to consider. We also need to consider the solubility equilibrium summarized by the solubility product constant $K_{sp}$. For example, for $\ce{CaCO3}$, $K_{sp} = 3.30\times 10^{-9}$ for the following equilibrium.

$$\ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-(aq)}$$


The equilibrium equation for neutralization is, with $K_b = 10^{-5}$:

$$\ce{CO3^2-(aq) + H2O(\ell) <=> HCO3^-(aq) + OH-(aq)}$$

$$K_b = \dfrac{[\ce{HCO3^-}][\ce{OH-}]}{[\ce{CO3^2-}]}$$

The overall reaction is then:

$$ \ce{CaCO3(s) + H2O (\ell) <=> Ca^2+(aq) + HCO3^-(aq) + OH-(aq)}$$

$$K = [\ce{Ca^2+}][\ce{HCO3^-}][\ce{OH-}] = K_{sp}K_b$$

Now we have what we need to evaluate the three bases thermodynamically:

$$\begin{array}{l|l|l|l|l} \text{Formula} & \text{solubility} & K_{sp} & K_b & K = K_{sp}K_b \\ \hline \ce{NaHCO3} & 96 \text{ g/L} & 1.31 & 2.24 \times 10^{-8} & 2.92\times 10^{-8} \\ \ce{CaCO3} & 1.3\times 10^{-2} \text{ g/L} & 3.30\times 10^{-9} & 1.00 \times 10^{-5} & 3.30\times 10^{-14} \\ \ce{Mg(OH)2} & 6.4 \times 10^{-6} \text{ g/L} & 5.61\times 10^{-12} & 50.1 & 2.81 \times 10^{-10} \\ \end{array}$$

From these data, the overall equilibrium constant is largest for $\ce{NaHCO3}$ because it has the highest solubility despite being the weakest base.

  • $\begingroup$ For the interested, lower case $k$, as in IUPAC here, about kinetics' rate constant. Which is, by convention, different to upper case $K$ about equilibrium constants, as in IUPAC here. $\endgroup$ – Buttonwood Mar 11 at 15:24
  • $\begingroup$ @BenNorris, if it isnt too much to ask, could you please provide me with the source of where you go the values displayed above on the table? $\endgroup$ – Shehab Al Dhobee Mar 11 at 20:15
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    $\begingroup$ All of these values are from the corresponding Wikipedia articles, (except the $K_{sp}$ of $\ce{NaHCO3}$, which I had to calculate from the solubility. $\endgroup$ – Ben Norris Mar 12 at 0:06

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