Question

pH be at the equivalence point? Volume NaOH added pH 0.39 The student collected the following data: Plot the data above in excel with pH values on the Y axis and the volume of NaOH added on the X axis Graphically determine the equivalence point and indicate the pH on the graph. b. 0.00 2.00 4.00 0.46 0.54 ii. 0.62 6.00 8.00 10.00 0.7 0.78 Show the following calculations on a separate sheet of paper: 0.87 12.00 0.96 1.07 14.00 16.00 18.00 Calculate the [H30+] Before adding NaOH Calculate the [H30*] After adding 10.00 mL NaOH Calculate the [H30+] 2 mL Before the equivalence point ii 1.19 iv. 1.35 20.00 22.00 24.00 1.56 V. 1.93 Calculate the [H30*] At the equivalence point Calculate the [H30] 2 mL past the equivalence point Calculate the [Ha0*] Volume after last addition vi. 24.50 2.09 25.00 25.50 vii. 2.35 3.06 iii. 26.00 11.4 26.50 11.8 27.00 28.00 12 12.2 29.00 12.3 3

Answer 1

(ii) [Hz ot] before adding Naott at p 0.39. ps -log[p] = -log!" 0.39 = mdog [130] 0.467384 - [ 130t - cantilog fo.39) (iv) After 10.00ml Nooth meane at pH = 0.78 = 0.165M. 0.78 = -log [HT] z = -log [lzot] =) [107] z antlog (-0.78) V 2nd, begle earivalenciansial mean at value g pH = 3.06. 3.06= -log [HTJ = dog [ 4307] - [ Bgst ] = antilog [ 3 ] = 0.00087M. نرمه له (اخ selen ( re 8 - الم ہے ا 0.00M 0.00M 8= -dog (ut] - log [ 48 ] : ) [267] vil) pay and part earwvalence poit de qt pH ally =) [113 ot] = dog (14) = OM. Niliy volume at last addition ie at: 29me pH = 123 ) [13 0 t] = log (-12 3] = OM.