Question

(5). Chapter 14

Calculate the pH of the solution produced when 20.0 ml of 0.125 M HCl are mixed with 25.0 ml of each of the following.

(a) 0.05 M Ba(OH)₂     (b) 0.12 M NH₃      (c) 0.08 M NaCl     (d) 0.20 M HClO₄    (e) water

Answer 1

5)

millimoles of HCl = 20 x 0.125 = 2.5

a)

millimoles of OH- = 25 x 2 x 0.05 = 2.5

here millimoles of acid = millimoles of base. so

pH = 7.00

b)

millimoles of NH3 = 25 x 0.12 = 3

NH3 + HCl   ----------------> NH4+Cl-

3         2.5                               0

0.5         0                               2.5

pOH = pKb + log [NH4Cl / NH3]

       = 4.74 + log [2.5 / 0.5]

       = 5.44

pH = 8.56

c)

millimoles of NaCl = 0.08 x 25 = 2

this is a salt it is no effect on pH

pH = -log 0.125

pH = 0.90

d)

this is strong acid . so

[H+] = 0.20

total [H+] = 0.125 + 0.20 = 0.325

pH = -log 0.325

pH = 0.488

e)

here adding water means diluted.

20 x 0.125 = 25 x M

M = 0.1

concentration of HCl = 0.1 M

pH = -log (0.1) = 1

pH = 1.00