# How to calculate the pH of a sodium bicarbonate-sodium hydroxide buffer solution?

This website claims that if you add $\pu{50 mL}$ of $\pu{0.05 M}$ sodium bicarbonate, and $\pu{5 mL}$ of $\pu{0.1 M}$ sodium hydroxide (and dilute to $\pu{100 mL}$), you should create a solution with $\mathrm{pH }= 9.6$.

I'm struggling a bit getting this result.

My attempt:

The sodium bicarbonate reaction would be:

$$\ce{NaHCO3 + H2O -> H2CO3 + OH- + Na+}$$

We know the $\mathrm{p}K_\mathrm a$ of $\ce{H2CO3}$ is 6.4, so the $\mathrm{p}K_\mathrm{b}$ of $\ce{HCO3-}$ is:

$$14 - 6.4 = 7.6$$

Furthermore, $\ce{pOH} = \mathrm{p}K_\mathrm b + \log\frac{[\ce{B+}]}{[\ce{BOH}]} = 7.6 + \log\frac{[\ce{H2CO3}]}{[\ce{HCO3-}]}$

However, how do we find $[\ce{H2CO3}]$ ?

I'm not entirely sure this is the right approach, but none of the other resources I've used to determine this have helped.

Your idea is good, but you have thought of the wrong reaction.

You have written that $\ce{HCO_3^-}$ hydrolyses to give $\ce{H_2CO_3}$ and $\ce{ OH^-}$, but think carefully when you add base, $\ce{OH^-}$ is in excess in this medium. So, shouldn't your proposed reaction proceed in backward direction instead? Moreover, your reaction is actually not going to occur.

When you are adding strong base, $\ce{HCO_3^-}$ acts an acid to give $\ce{CO_3^2-}$ like:

$$\ce{NaHCO_3 + NaOH -> Na_2CO_3 +H_2O}$$ or, more simply $$\ce{HCO_3^- + OH^- -> CO_3^2- + H_2O}$$

Initially, number of millimoles of $\ce{HCO_3^-}$ were $=50\times0.05 = 2.5$ and number of millimoles of base added $=5\times0.1 = 0.5$

So, $\ce{CO_3^2-}$ produced will be also 0.5 millimoles as the base added was a limiting reagent, and $\ce{HCO_3^-}$ left $= 2.5-0.5 = 2$ millimoles.

As, you can see now, the solution acts an acid buffer as, $\ce{HCO_3^-}$ is a weak acid and $\ce{CO_3^2-}$ is the salt after reacting with a strong base. According to Henderson-Haselbach equation,

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log{[\text{salt}]}/[\text{acid}]$$ $\mathrm{p}K_\mathrm{a}$ of $\ce{HCO_3^-}=\mathrm{p}K_{\mathrm{a2}}$ of $\ce{H_2CO_3^}=10.3$ . Now [$\ce{HCO_3^-}$] = $\pu{0.02 M}$, and [$\ce{CO_3^2-}$] = $\pu{0.005 M}$. Now if you calculate $$\mathrm{pH}= 10.3 +\log\frac{0.5}2 = 10.3- \log(4) = 9.69$$ So, the final $\ce{pH}$ of the medium has come out to be $= 9.69$.

The difference in digits after decimal may be due to some given $\mathrm{p}K_\mathrm{a2}$ which is experimental and you should do calculation according to that.