Consider the titration of 39.5 mL of 0.201 M HCl with 0.123 M KOH. Calculate pH...

Question

Consider the titration of 39.5 mL of 0.201 M HCl with 0.123 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. (Assume that all solutions are at 25°C.)

So far this is what I have. For d and e I keep on getting negative molarity and I am unsure on whether or not I got the initial pH right. Please show your steps.

(a)    0.0 mL

#mols of HCl = molarity x volume in L

=0.201 x (39.5/1000) = .0079395

.0079395/.0395L=.201M

=-log(.201)=.6968

(b)    10.0 mL

pH=.868

(c)    40.0 mL

pH=1.42

(d)    80.0 mL

0.123(80.0/1000)=.00984

# mols left = .00793-.00984=-.0019005??

(e)    100.0 mL

0.123(100.0/1000)=.0123

# mols left = .00793-.0123= -.0044??

Answer 1

39.5 ml of 0.201 M HCl

a) no KOH added: You have calculated the correct pH

b) 10 ml of 0.123 M KOH:

0.010 L x 0.123 mol/L = 0.00123 mol KOH

0.0395 L x 0.201 mol/L = 0.00794 mol HCl

0.00794 mol - 0.00123 mol = 0.00671 mol HCl remaining

Final volume = 10 ml + 39.5 ml = 49.5 ml = 0.0495 L

Final [HCl] = 0.00671 mol/0.0495 L = 0.136 M

pH = -log 0.136 = 0.868

c) 40 ml of 0.123 M KOH:

0.040 L x 0.123 = 0.00492 moles KOH

0.00794 mol - 0.00492 mol = 0.00302 mol HCl remaining

Final volume = 79.5 ml = 0.0795 L

Final [HCl] = 0.00302 mol/0.0795 L = 0.0380 M

pH = 1.420

d) 80 ml of 0.123 M KOH:

0.080 L x 0.123 mol/L = 0.00984

0.00984 - 0.00794 = 0.0019 moles KOH remaining (NOTE: KOH is in excess, no longer HCl in excess)

Final volume = 119.5 ml = 0.1195 L

Final [KOH] = 0.0019/0.1195 L = 0.0159 M

pOH = -log 0.0159 = 1.799

pH = 14 - 1.799 = 12.201

e) 100 ml of 0.123 M KOH:

0.1 L x 0.123 mol/L = 0.0123 mol KOH

0.0123 - 0.00794 = 0.00436 moles KOH remaining (NOTE: KOH is now in excess, no longer HCl)

Final volume = 139.5 ml = 0.1395 L

Final [KOH] = 0.00436 mol/0.1395 L = 0.0313

pOH = -log 0.0313 = 1.505

pH = 14 - 1.505 = 12.495