Question

Calculate the pH for each case in the titration of 50.0 mL of 0.120 M HCIO(aq) with 0.120 M KOH(aq). Use the ionization constant for HCIO. What is the pH before addition of any KOH? pH = What is the pH after addition of 25.0 mL KOH? pH = What is the pH after addition of 35.0 mL KOH? pH = What is the pH after addition of 50.0 mL KOH? pH = What is the pH after addition of 60.0 mL KOH? pH =

Answer 1

PH = -log fu+] [PH=4.16 After addin of 25 mL of 0.12M KOH - 2 25ml x 612mol = 3 mmol base Helot Koh RocI+H₂O and Bare Salt • 1 and Rase reacts with ind aid to foron 1 mol salt .: 3 mmol bid react with 3 mond aid to produce 3 mmol Salt Asalt - 3 mmol V total = 50ml + 25m2 = 75 m2 Canid) (Base) (salt) = Asalt = 3nmol = 0.00M Vtotal 75 ph L (Acid] = nacid = (6 - 3 mmol = 0.04M Vertal initial Reacted 75mL pth pka + log/salt] =pka + log/0.04 ) pH = pka + 1 pH = 7.3961 - 3) After 35ml of 0.12 mol Koh 4.2 mmot hare nsalt = 4.2mmal [salt] = 4.2 mmol = 0.05M (50+35 mL (Acid ] = (6-4.2) ontmol = 0.02118M 50 +35) mL pH = 7.398 +log (9:05) pH = 7.766
2
7 After addition of some of 0.121 Bar - 6 mmol Bar Sathy - ohmol = Salt) = 6ohmal - 0.06M. (50+50 ML acid lax all acid reacted with all house I only salt equivalence point : Kollis a salt of weakoud Strong Bare pH = p + 1 [pka + lug [] rc. Loner +7+47.398+layloco) (pH = 10.091 5) after some of 0.12 mol KOH 1 7.2 mmol Bax or 6 mmol pare Emmol Salt excess Bar = 7.2mmd - 6 mmol = 1.2 nmol [Base] = 1,2 mmol = 0.010am (50+ Goom pOH = -log fort] = log (6.0log) POH = 1.96 : PH= 14 POH TpH = 12.038 Thus, pH after excess addition of Base is 12.030
3

This is FIRST PAGE

50mL of 0-12 M HClO 50ml x 0 12 mol = 6 mmol Holo t-o C Ka (H(16) = 4x168 pka= -logka = 7.398 Before Addin Kom Holo H+ + oci- eam C-cd cd ca. C- Conc. of HClO = 0.12 M. & degree of dissociation Ka= [ht] Tour] =(cd) (Cd) = ca- Chcloc -cd ind Ka=cd² 1 de 1 d=5.794154 [+] = Cd = 0.12) (5. 77 804) = 6.93x10 m