Question

47. What is the shape of the titration curve (plot of pH vs. mL of added base) for titration of a weak acid by a strong base? In a titration, 10 mL of 0.10 M HF was titrated with 0.20 M KOH. Calculate the pH of the resulting solution after the addition of the following volumes of base: 0 mL, 2.5 mL, 5.0 mL, and 6.0mL. What volume of base is needed to reach the equivalence point? What species is present at the equivalence point and what is the pH at the equivalence point?

Answer 1

hationof +itr aton corve o Tue aci d by ueak a strong bwae aguivalamee poink PH NM for Hf Pkaz 3. when Nao H added there is H301 HF Kaz 20 -4 G6X10 O'0-20 -66x104+6Xlo4 46X105 2Y = 7-8 X10-3 LH30 7.X lo3 pH= -103 (7-9x la) =62.10 Scanned with CS CamScanner

2.5 ml Naoltis asdod Natf- 2.5X0.2 20-04 1042.5 Ox01- 2.5 X0.2 [NF] e0.04 1042.5 pH= CNatf1 pka+ log H O04 3.1 10 ren 5mi Nao His ablad 067 104 5 CHEIE 16x0.1-5X0 2 10 +5 2 Aftar asgi Hm of 5m Naot Haot Tuu- NOw f uis aeak aci & Hf O4 Ka 66X10 k - l57X10 1 66X 10-4 DHE7LON - I 1-SIX10 11 o.O67 LOM 0075 X10 pOH 10 0075 X10 = 5.44 G7 p/t 14- 8.0032 5.99 67- CS Scanned with CamScanner

AB.&IHon of 6m Naot C6-5M0.2 Atter O0125 M. 10+6 pH-109(0r0125) 03 PH14-1.90312 09 bo of NaoH is 5 mi vouune Now egumed to neach equival onee point OH is precent cjn gcvalame pomtie baoic pH 8-0032 09 Scanned with CS CamScanner