{"id":5064,"date":"2012-05-17T16:28:00","date_gmt":"2012-05-17T16:28:00","guid":{"rendered":"https:\/\/www.masterorganicchemistry.com\/?p=5064"},"modified":"2025-09-30T13:31:26","modified_gmt":"2025-09-30T18:31:26","slug":"a-handy-rule-of-thumb-for-acid-base-reactions","status":"publish","type":"post","link":"https:\/\/www.masterorganicchemistry.com\/2012\/05\/17\/a-handy-rule-of-thumb-for-acid-base-reactions\/","title":{"rendered":"A Handy Rule of Thumb for Acid-Base Reactions"},"content":{"rendered":"

Reversible And Irreversible Acid-Base Reactions <\/strong><\/p>\n

Last time we learned about pKa<\/sub><\/a> and how it’s the closest thing we have to a universal measurement of \u00a0the strengths of all kinds of different acids and bases.<\/p>\n

I also referred to a post on how to use a pKa<\/sub> table<\/a> (key lesson: stronger acid plus stronger base gives weaker acid and weaker base).<\/p>\n

We can use pKa<\/sub> to estimate the equilibrium constants for acid-base reactions.<\/p>\n

A handy rule of thumb for an acid-base reaction is that if the\u00a0acid\u00a0<\/strong> and the\u00a0conjugate acid<\/strong> are separated by more than 10 pKa<\/sub> units, the reaction is essentially irreversible<\/strong>.<\/p>\n

\"summary-rule<\/a><\/p>\n

Table of Contents<\/strong><\/p>\n

    \n
  1. An Irreversible Acid-Base Reaction: Strong Acid (HCl) Plus Strong Base (NaOH) Giving Water<\/a><\/li>\n
  2. An Easily Reversible Acid-Base Reaction: Methanol (pKa 15.2) With Water (pKa 14)<\/a><\/li>\n
  3. At What Point Does An Acid-Base Reaction Become Irreversible For Practical Purposes?<\/a><\/li>\n
  4. Application: In The Claisen Condensation, The Starting Ester Is Less Acidic Than CH3<\/sub>OH By About 10 pKa<\/sub> Units<\/a><\/li>\n
  5. Notes<\/a><\/li>\n<\/ol>\n
    \n

    <\/a>1. An Irreversible Acid-Base Reaction: Strong Acid (HCl) Plus Strong Base (NaOH) Giving Water<\/h2>\n

    On one extreme<\/strong>, we have one mole of a really strong acid – let’s say hydrochloric aid (HCl), pKa<\/sub> \u20138. And to it, we add (slowly!) a solution of water containing one mole of sodium hydroxide (the conjugate base of water, pKa<\/sub> 14).<\/p>\n

    \"-irreversible<\/a><\/p>\n

    HCl and NaOH react to give water and NaCl . How favorable is this reaction? We can make a rough estimate. The pKa<\/sub> of HCl is -8. Sodium hydroxide is the conjugate base of H2<\/sub>O (pKa<\/sub> 14).<\/p>\n

    That’s a difference of about 22 pKa<\/sub> units – and since each pKa<\/sub> unit represents one order of magnitude, this reaction is favorable with an equilibrium constant of about 10 to the power of 22.<\/p>\n

    For all intents and purposes, a reaction with an equilibrium constant this huge is irreversible<\/strong>.<\/p>\n

    That is to say that HCl and NaOH are completely consumed when they react together, giving only H2<\/sub>O and NaCl.<\/p>\n

    <\/a>2. An Easily Reversible Acid-Base Reaction: Methanol (pKa 15.2) With Water (pKa 14)\u00a0\u00a0<\/strong><\/h2>\n

    What about the other extreme:<\/strong> the reaction of methanol (pKa<\/sub> of 15.2) with sodium hydroxide (the conjugate base of water, pKa<\/sub> 14)?<\/p>\n

    \"reversible<\/a><\/p>\n

    Neither side of the acid-base reaction is strongly favored. Here we’re dealing with a very small difference in pKa<\/sub> – only 1.2 pKa<\/sub> units.<\/p>\n

    So the equilibrium constant here would only be about 10 to the power of 1.2 —> 15.8\u00a0 toward giving the weaker acid (CH3<\/sub>OH) and the weaker base HO(\u2013).<\/p>\n

    At equilibrium we’d expect to have a mixture of about 94% HO(-) [the weaker base] and 6% H3<\/sub>CO(-)\u00a0 [the stronger base].<\/p>\n

    In other words, both species<\/strong> are present in solution.<\/p>\n

    <\/a>3. At What Point Does An Acid-Base Reaction Become Irreversible For Practical Purposes?<\/h2>\n

    \u00a0So how far can we stretch this?<\/strong> In between these two extremes, at what point does\u00a0 a reaction become irreversible for practical purposes?<\/p>\n

    There’s no hard and fast rule on this. But for practical purposes, a good rule of thumb is about 10 pKa<\/sub> units.\u00a0<\/strong><\/p>\n

    That is to say, if the difference in pKa<\/sub>‘s between an acid and a base (actually,\u00a0the conjugate acid of the base<\/em><\/span>)\u00a0 is about 10 pKa<\/sub> units or less, it is useful to consider their acid-base reaction to be in equilibrium.<\/p>\n

    Think about what that means – a ratio of one molecule in 10 billion can make the difference in a reaction!<\/p>\n

    One in 10 billion might not sound like a lot. But when you consider that a mole contains 10 to the power of 23 molecules, and each of them are colliding millions of times per second, the odds aren’t really as bad as they look.<\/p>\n

    f your only chance of buying a private jet rested on you winning a Powerball lottery – but you were able to fill out hundreds of thousands of entries per second, every second, you’d be at the G5<\/a> dealer by next Tuesday.<\/p>\n

    Here’s an example you’ll see in Org 2. The Claisen condensation begins with the deprotonation of an ester (pKa<\/sub> ~24) by an alkoxide ion (conjugate base of an alcohol, pKa<\/sub> ~15). (See article:The Claisen Condensation<\/a><\/em><\/span>)<\/p>\n

    That’s disfavored by about 10 to the power of 9, since we’re going from a weaker base (alkoxide) to a stronger base (deprotonated ester, a.k.a ester enolate).<\/p>\n

    \"a<\/a><\/p>\n

    Even though there’s only a small amount of the deprotonated ester present at equilibrium, this can be enough<\/strong> to get the reaction to go! You can take my word for it – that this rule of thumb applies \u00a0– and leave it there.<\/p>\n

    Or if you’d prefer to go through an actual application of this concept, I’ll finish up with that.<\/p>\n

    <\/a>4. Application: In The Claisen Condensation, The Starting Ester Is Less Acidic Than CH3<\/sub>OH By About 10 pKa<\/sub> Units<\/strong><\/h2>\n

    One application of this concept can be found in the Claisen condensation of esters. The Claisen condensation involves the addition of a deprotonated ester (an “enolate”) to another equivalent of an ester, through an addition-elimination reaction (see article: The Claisen Condensation<\/a><\/span>)<\/em><\/p>\n

    The first step is deprotonation of the ester by an alkoxide ion [in this case CH3<\/sub>O(-) ] as \u00a0mentioned above. This enolate can then attack a second equivalent of ester, which then eliminates an equivalent of alkoxide ion.<\/p>\n

    This reaction is also potentially reversible. However, the protons of the new product – a “beta-keto ester” – are considerably more acidic than those of the starting ester, and an acid-base reaction between it (pKa<\/sub> 12) and alkoxide (pKa<\/sub> 15) is quite favorable.<\/p>\n

    \"applicaiton<\/a><\/p>\n

    The equilibrium eventually favors the final product, because the conjugate base of the beta-keto ester (pKa<\/sub> 12) is considerably weaker than methanol (pKa<\/sub> ~15).<\/p>\n

    In other words, even though the first step is extremely disfavored, this is made up for by the fact that there is a very good “driving force” for the subsequent reaction.<\/p>\n

    Next Post: Acid-Base Reactions Are Fast\u00a0<\/a><\/p>\n

    \n

    <\/a>Notes<\/h2>\n

    Related Articles<\/strong><\/p>