Question

E. Titration calculations 1. A 25.0-ml sample of 0.100 M HCl is titrated with 0.125 M NaOH. How many milliliters of the titrant will be need to reach the equivalence point? 2. A 25.0-ml sample of 0.100 M Ba(OH)2 is titrated with 0.125 M HCI. How many milliliters of the titrant will be need to reach the equivalence point? 3. For the following titrations, determine if the equivalence points will be acidic, basic, or neutral i. NH3 titrated with HCI ii. Ba(OH)2 titrated with HCI iii. HF titrated with NaOH

Answer 1

ų < : balanced equation is Nach + H2O (a) e Het Naoth (aw) (ar) - we know that ni nz Mi= rolarity of Hel= 0.1007 = volume of Hu= 25 ML ni = Number of roles of Haz 1 Mq = Molarity of Naolt=0.125 V = volume of Naut= ? ML roles of Nault=1 n2 = Number of . hov, n - Hz V, 1, Vq = Minna = Meni 0.100 XP5X/ 0.125x1 20 ML of vaoh 12/12 int - = i volune gori

® balanced equation Ba(OH)e + 2HCl lag) (291) - + 2H₂O Back aw) we know that - Mive ni Meu ho Mia Molarity of Bacouble = 0.1007 Vi = Vol Une of Balottle = 2514 Maar nia Number of roles of BaCOH/2 = 1 Mq = Molarity of HU = 0.251 Ve = vo lune of Hel=?he ne = Number of Moleg of Ital= { .: MW M2 = Mevani . . Vq = milim roni = 0.100 x 25x2 0.125 XI = 40 ML vosure of Ha 13/13 rt) vej a Home

(3) i NH3 titrated with Hi Ang acidic solution - Ultz is weak base, He is strong acid. Therefore Nurber. of to tong are more. So the Solution is acidic (ii) Balottle titrated with He And Neutral solution, Ba(OH)2 Strong base, he is strong acid (ii) HF titrated with NAOH Ang basic solution- - HE is weak acid, Paolt is strong buse. there are number of ott Iong are more. so the solution is basic.