Question

B. DETERMINATION OF THE n pK, OF AN UNKNOWN ACID Wnw TABLE 8.2 2nd Determination 3rd Determination 1st Determination pH mL of NaOH PH mL NaOH mL NaOH pH B.0 303 2.9 3 19 .42 (0 2 -awa 3.00 727 42 41 166 1210 thl 402 426 | 213 39 4.46 210 .42 113 9 1210.a 13 442 11.4711 1441 11 4 ly 9 I125 102 1G5 115 TABLE 8.3 1st Determination 2nd Determination 3rd Determination volume at equivalence point volume at half equivalence point K, (acid equilibrium constant) average standard deviation

Answer 1

To solve this problem first we are to make graph of pH v/s Volume of NaOH. From this graph we can find out volume of NaOH at equivalence point and volume of NaOH at half equivalence point (mid point) using the following procedure: (Note- In following procedure two equivalence points are shown but your data has only one equivalent point and one half equivalent point)

value of Ka can be calculate using following formula:

Unknown Acid delta mlL (delta pH)/ (delta First derivative vs volume of NaOH NaOH NaOH delta pH LNaOH) 1.99 2.07008 0.04 0.065 0.085 0.08 0.08 0.07 2 1. 4 Seoond equivalence point 2.370.17 2.53 0.16 2.69 0.16 2.83 0.14 5.79 5.79 6.63 .97 7.22 0.25 7.44 0.22 7.6G 0.22 7.93 0.27 8.34 9.631.29 11.33 1.7 11.80.47 11.9Z 0.12 12.240.32 12.340.1 12.45 0.11 12.540.09 12.62 0.08 12.690.07 10 12 2.96 18 20 0.84 0.34 0.42 0.17 0.125 0.11 0.11 0.135 0.205 0.645 0.85 0.235 0.06 0.16 0.05 0.055 0.045 0.04 0.035 Volume of NaOH (mL] Volume of NaOH mL) 14 mL First equivalerce point pH can be found againstat midpoint(? mL) 24 Midpoint Vol. 7 mL 28 30 Here two equivalence points and two half equivalence points (midpoint) are present then the pKa will also be two. First pKa and second pKa 0.41 In the plots it is may be a dibasic acid. whea weak dibasic acid is titrated, there are two equivalence points will be present First equivalence point will occur when H+ is dissociated. And second equivalence points will come when second H atom that 36 38 removed acid molecule The half equivalence point midpont), also indicated in the figure, is when the number of moles of stuong base added eqals 114 of the moles of the weak acid hat are present. For this reason, the midpoint is half of the equivalence point. Notice that there are two midpoints as there are two equivalence points. At the midpoint, pH equals the value of pKa because there is 5D:50 mixture of the weak acid and the shong base. To quantifythis, the Henderson-Hasselbalch Approximation can be used: 50 pI pKa logpH pKa when(loglA/HA 1 at 50:50 mole ratio of acid and base)

pka = -log ka (by taking antilog ka can be find out)