# Chemistry Question about Acids and Bases

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It is about acids and bases and half tit rations. It is a calculation found in an A level Chemistry Past Paper from Malta.

 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #11 Sep 18, 2011, 10:50 AM

Quote:
 The pH of a solution obtained by addition of 8.2 cm^3 of sodium hydroxide to 25.0 cm^3 of the acid is 4.2. Calculate the concentration of hydroxonium ions in this solution and hence find the dissociation constant of the acid. (5 marks) Moles of acid in 25cm^3 = 5.08 x 10^-4 Concentration of Sodium Hydroxide = 0.0310 Below is the answer given by my tutor: From the previous calculation, the number of moles of monoprotic, weak acid in 25.0cm^3 of solution is 5.08x10^-4 moles. The number of moles of sodium hydroxide in 8.2 cm^3 of sodium hydroxide can be calculated too. Moles NaOH = volume x concentration = (8.2/1000) x 0.0310 = 2.54 x x10^-4 moles Since each mole of sodium hydroxide reacts with one mole of the monoprotic acid, then 2.54 x 10^-4 moles of the base react with 2.54 x 10^-4 moles of the acid. This forms 2.54 x 10^-4 moles of salt Na^+A^-. The number of moles of acid left is the difference between the original number of acid and the moles which reacted with the base. The 2.54 x 10^-4 moles of salt NaA are completely dissociated in aqueous solution forming the same number of moles of A^-, while the weak acid HA dissociated. Since the number of moles of HA and A^- is the same, the concentration of these two species is equal too. The [H^+] can be found from the pH. pH = - log [H^+] [H^+] = 10^(-pH) [H^+] = 10^(-4.2) [H^+] = 6.31 x 10^-5 mol dm^-3 The equation for the dissociation constant for the acid, Ka, is: $K_a = \frac{[H^+][A^-]}{[HA]}$ * Since the number of moles of HA and A^- is the same, the concentrations of these two species cancel out of the equation. Ka = [H^+] Substituting the [H^+], then: Ka = 6.31 x 10^-5 mol dm^-3 *That's where I don't agree. I think that the concentration of A^- should not be taken the same as that of HA because there are 2.54 x 10^-4 moles of A^- coming from the salt but also 6.31 x 10^-5 from the unreacted acid. Am I Correct in my thinking of what?
Well, if you had written the equations, you wouldn't have had any doubt.

The acid is monoprotic, as the base is monobasic. The mole ratio is therefore 1:1. Since there is twice the amount of acid, there is 2.54 x 10^-4 moles of acid in excess.

So, the equations for the reactions occuring:
$NaOH + HA \rightarrow NaA + H_2O$

Now, to the equilibrium equations:
$NaA \rightarrow Na^+ + A^-$

$HA \rightleftharpoons H^+ + A^-$

There is initially 2.54 x 10^-4 moles of HA, and there is a common ion effect for the A^- ion. The equilibrium constant equation thus become:

$K_a = \frac{[H^+][A^-]}{[HA]}$

$K_a = \frac{$6.31\times10^{-5}$$\frac{1000$$2.54\times 10^{-4}+x$$}{33.2}$}{$\frac{1000$$2.54\times 10^{-4} - x$$}{33.2}$}$

x here is the amount of acid which has dissociated. It's the number of moles (2.54x10^-4) added to x to get the total number of moles of A^- and it is the total number of moles of HA (2.54x10^-4) minus the amount of moles x of HA that dissociated.

Since HA is a weak acid, it is understood that x is very very small and for ease of students, we take [A^-] = [HA]

Also, unless you know Ka, you would not be able to find x here, remember that the equation deals with concentration, which is the number of moles, divided by the volume times the total volume

For instance, with a Ka of 6.31x10^-6 mol dm^-3, you will have something of the other of 3.7x10-5 mol from the 2.54x10^-4 mol of HA, which is ten times less than the original content. That margin of error is okay in A-level and does not require you to go into very precise mathematical skills only to get a value slightly different.