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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.

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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.

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