Acid Base Practice Problems
Calculating the Ka of an Acid Given Concentration and pH
The pH of a 0.3 M ethanoic acid (HC2H3O2) is found to be 2.63. What is the Ka of ethanoic acid?
- Write the Ka expression for the dissociation of ethanoic acid:
\(\ce{CH3COOH_{(aq)} <=> CH3COO-_{(aq)} + H+_{(aq)}}\)
\(\ce{Ka = \frac{[CH3COO^-_{(aq)}][H+_{(aq)}]}{[CH3COOH_{(aq)}]}}\)
-
Construct an ICE table for dissociation of ethanoic acid:
Phase |
\(\ce{CH3COOH}\) |
\(\ce{H+}\) |
\(\ce{CH3COO-}\) |
Initial (M) |
0.3 |
0 |
0 |
Change (M) |
-x |
+x |
+x |
Equilibrium (M) |
0.3 - x |
x |
x |
Substituting these values into the Ka expression we get:
\( K_a = \frac{{[\ce{X}][\ce{X}]}}{{[0.3]}} \)
Since 0.3 is significantly greater than x (this is a weak acid, so the extent of deprotonation (concentration [H+] at equilibrium) is negligile compared to the intial concentration of ethanoic acid), we can ignore this term in the denominator.
-
Determine the concentration of H+ at equilibrium given the pH:
\([\ce{H+}] = 10^{-\text{pH}}\)
\([\ce{H+}] = 10^{-2.63} = 0.0023\)
-
The concentration of H+ is the value of x in our Ka expression. Substitute to solve for Ka:
\( K_a = \frac{{(0.0023)(0.0023)}}{{0.3}} = 1.8 \times 10^{-5} \)
Calculating the Ka of an Acid Given Tiration Curve
Given titration data, how can one determine the Ka for an unkown acid?
At the equilavence point of a titration, the analyte (solution whose quantity or concentration is being determined) has fully reacted with the analyte. In other words, at this point, moles of acid = moles of base. In a strong acid strong base titratioh, the pH at teh equivalence point is neutral, as the species that exist are a netural salt and water. For example, consider the titration of HCl with NaOH. At the equivalence point, just product exists according to the following reaction:
\(\ce{HCl + NaOH -> NaCl + H2O}\)
Since NaCl does not have acidic or basic properties, the pH at this point is neutral (7). What about a weak acid strong base titration, such as the tiration of acetic acid with NaOH? In this titration, the pH at the equivalence point will be greater than 7, as the species present will be:
\(\ce{CH3COOH + NaOH -> CH3COONa + H2O}\)
Sodium acetate is a strong electrolyte and will dissociate into CH3COO
- and Na
+. The pH at the equivalence point is determined by the base, CH3COO
-, and is greater than 7.
What if one did not know the Ka for acetic acid but had the tiration data? In a weak acid strong base titration (or weak base strong acid titration), a buffer region exists between the start of the titration and the equivalence point due to the present of a weak acid base conjugate pair. In the above example, this would be acetic acid (CH3COOH) and its conjugate base, acetate (CH3COO
-). The pH of a buffer solution can be determined by the Henderson-Hasselbalch equation.
\(\ce{pH = pKa + log10(\frac{[A-]}{[HA]})}\)
When the concentration of base and acid are equal, the pH is equal to the pKa. In the titration, this is known as the
half-equivalence point, and occurs at half the volume of titrant added relative to the equivalence point. By noting the pH at which the half-equivalence point occurred during the titration, one then knowns the pKa of the acid. Consider the curve below for the titration of acetic acid with NaOH:
Here, we see that the volume of NaOH added at the equivalence point is 50 mL, and therefore 25 mL at the half-equivalence point. The pH at half-equivalence point is 4.767. Thus, the pKa of acetic acid must be 4.767. We can then find the Ka:
\(\ce{Ka = 10^{-pKa} = 10^{-4.767} = 1.7 \times 10^{-5}}\)
The Ka of acetic acid is 1.7 x 10
-5.