Question

You titrate 50.0 mL of 0.400 M HClO₄ (a strong acid) with 0.400 M NaOH.

a. Write the balanced equation.

b. What is the pH at the beginning of the titration (0.0 mL of NaOH)?

c. What is the pH after adding 30.0 mL of NaOH?

d. At the equivalence point, is the pH greater than, less than, or equal to 7.00? Explain briefly.

Answer 1

The answer is attached.

a) gation) for the reaction Balanced reaction Cequation) for t between Helou and Nooh is : Helon at NaOH Naciou & H2O Since in this . - equation, the number of equation 'H', 'ci , 'o' and 'Na' anno Scume on ". This is the balanced chemical equation. on both side Na some are b) pH when nach is not present (o.omu). Since this is a strong acid | strong base titration therefore same amount of it (from recoy) reacts with oh (from Naote) to give H₂O Het tot ² H₂O. At the beginning, moles of the - volume (in 'L) & concentration of het (inm) s 50 X10-3LX 0.400 m = 0.02 moles Limilarly, No. of moles of Oh OLX0.400m 2 O moles. : concentration of it in the total volume (somL) in 50 x103 L ...pr = -log[ht] = -log (0.4) = = 0.02 mol = 0.4 m = [ht] TpH = 6 3 8 |

les of c) After adding 30 mL of Naoth, no. of On (NaOH) = (30x18-3) L x 0.40om = 0.012 mor No. moles of Ht - 50x10 -0.02 mol of Ht a sox10-3 Lx0.4oom het of remaining concentration of the = 0.02 -0.012 = 8x10-s mol [HA] = 8x10-3 mol = 8x10-3 = 1x1022M (50+30)x1031 pH = -log [H+] = -log Cix 10-2) = 2 . &xion . ) At equivalence point, equal number of moles of relay and Nadh will react as this is a strong acid I strong base titration. :.0.02 mol of Heroy will reach equivalence - point when 0.02 mol of Noote will be added from wisi = Vasz Hao X Shady = 'Naon & Snoot =) 50mL X 0.4oom = Hash xoyoom = 'Noon = 50mL : remaining moles of from Hclou Het = 0.02 -0.02 ao moles

concentration of tt - remaining me aining moles of the Helou) Total volume o moles aomart Csom 50 ) X20-31 (50+50X1003 So, the concentration of it from dissociation of water will only be taken into account.. Dissociation constant of water. = 1X18-7 pH = -log (1x1077) = 7. . At equivalence point, the ph is equal to 7 since at equivalence point all holou will be neutralized from North and only [et] from water will be considered