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| | From:  goldie647 (Original Message) | Sent: 3/14/2008 3:16 PM |
In the following question how do you know that you have to use the Henderson Hasselbalch equation?
Calculate the percent ionization of 0.10M butanoic acid
(Ka = 1.5 x 10^-5) in a solution containing 0.050M sodium butanoate.
Because my first thought was to just use:
%ionization = (H+ / Molarity) x 100
Any suggestions
Thanks
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| | | From: ·Steve· | Sent: 3/14/2008 9:13 PM |
The Henderson-Hasselbalch equation can only be used when you have a conjugate pair in the same solution. That is, a weak acid plus its conjugate base, or a weak base plus its conjugate acid. These are buffer solutions by definition. If you do not have both present, you will either be dividing by zero or taking the log of zero in the equation! Percent ionization... That is right. Calculate H+ concentration if you have a weak acid solution, as you do here, and divide by the total concentration of the weak acid and multiply by 100. That will be the percentage of the weak acid that is in ionized, or dissociated, form. So, in this problem, you can calculate the pH from the Henderson-Hasselbalch equation and get the H+ concentration from that. The percent ionization will be [H+] / 0.10 M times 100. Steve |
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I know that butanoic acid is a weak acid, but how can you tell that sodium butanoate is its conjugate base?
I know that for a conjugate base you remove a H+ ion.
More confused |
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| | | From: ·Steve· | Sent: 3/14/2008 11:08 PM |
That's right, when you remove H+ from butanoic acid, HC4H7O2, you get the butanoate ion, C4H7O2�?/SUP>. The sodium ion is a spectator as far as pH is concerned, since it has no effect on H+ or OH�?/SUP> concentration. When you react an acid with a base, the conjugate base of the acid is formed: HC4H7O2 (aq) + NaOH (aq) ––�?gt; NaC4H7O2 (aq) + H2O (l) Or in net ionic form, HC4H7O2 (aq) + OH�?/SUP> (aq) ––�?gt; C4H7O2�?/SUP> (aq) + H2O (l) weak acid conjugate base of weak acid |
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I guess I am confused because I do not know the formula for these two substances. So how could I know how to setup the reaction to see that sodium butanoate was the conjugate base of the weak acid butanoic acid.
Thanks
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| | | From: ·Steve· | Sent: 3/17/2008 1:46 AM |
You actually do not need to know the formulas, we can use "generic" ones. Dissociation of a weak acid (initial concentration = "I"): HA (aq)  H+ (aq) + A�?/SUP> (aq) Initially: I 0 0 At Equilib: I �?x x x Hydrolysis of a weak base (initial concentration = "I"): B (aq) + H2O (l) HB+ (aq) + OH�?/SUP> (aq) Initially: I �?nbsp; 0 0 At Equilib: I �?x �?nbsp; x x If you start with a negative ion weak base, we can write almost the same thing: B�?/SUP> (aq) + H2O (l) HB (aq) + OH�?/SUP> (aq) Initially: I �?nbsp; 0 0 At Equilib: I �?x �?nbsp; x x In other words, even without the actual formulas, the calculation of x is the same.  Steve |
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