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I read somewhere that when $\pu{100 mmol}\ \ce{NaHCO3}$ are mixed with $\pu{50 mmol}\ \ce{HCl}$, a buffer arises. I'm confused, since $\ce{HCl}$ is a strong acid, and I thought the classic buffer is weak acid, weak conjugate base in equal amounts.

Is it the fact that the strong acid is added in only half the amount as the weak base that creates the buffer?

Or is it that the $\pu{50 mmol}\ \ce{HCl}$ will turn half of the $\pu{100 mmol}\ \ce{NaHCO3}$ into $\pu{50 mmol}\ \ce{H2CO3}$ which then with the remaining $\pu{50 mmol}\ \ce{NaHCO3}$ comprises a classic buffer?

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    $\begingroup$ I'd very much like a reason for the down vote? This is my first question here, so honestly don't know this question constitutes a "bad question". $\endgroup$
    – chsimmons1
    Commented Jan 10 at 9:47
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    $\begingroup$ Formatting guides for texts and formulas/equations/expressions. // Downvoting is anonymous and some users do not provide feedback. // Review the guide How to ask and Asking FAQs to prevent clarification requests, objections, downvoting or closure. $\endgroup$
    – Poutnik
    Commented Jan 10 at 9:54

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Your second idea is right, the acid partially neutralizes bicarbonate.

$$\ce{HCO3-(aq) + H3O+(aq) -> H2CO3(aq) + H2O}$$

It is effectively the buffer $\ce{HCO3-(aq)/CO2(aq)}$, as $\ce{H2CO3(aq)}$ is about just one percent of total dissolved $\ce{CO2}$.

The problem is such a buffer is unstable because there is not usually established the equilibrium between aerial and dissolved $\ce{CO2}$ and it therefore gradually escapes from the solution, with pH drifting to higher values.

$$\ce{H2CO3(aq) <=> CO2(aq) + H2O <=> CO2(g) + H2O}$$

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