Question

Calculate the pH of each of the following strong acid solutions.

(a) 0.00458 M HCl

pH =

(b) 0.643 g of HClO4 in 37.0 L of solution

pH =

(c) 60.0 mL of 1.00 M HCl diluted to 4.10 L

pH =

(d) a mixture formed by adding 69.0 mL of 0.000760 M HCl to 46.0 mL of 0.00882 M HClO4

pH =

Answer 1

a) 0.00458 M HCl

pH = -log[H+]

PH = -log[0.00458]

Ph = 2.34

(b) 0.643 g of HClO4 in 37.0 L of solution

no. of moles = 0.643/37 = 0.02

molarity = (wt/mol.wt)*(1/vol. in ml)

M = (0.643/100.5)*(1/37)

M = 1.73*10^-4 M

Ph = -log[ 1.73*10^-4] = 3.76

(c) 60.0 mL of 1.00 M HCl diluted to 4.10 L

M1V1 = M2V2

1*60 = M2*4100

M2 = 60/4100

M2 = 0.015 M

Ph = -log[0.015]

Ph = 1.83

(d) a mixture formed by adding 69.0 mL of 0.000760 M HCl to 46.0 mL of 0.00882 M HClO4

M = [M1V1 + M2V2]/(V1+V2)

M =[0.00076*69 +0.00882*46]/(69+46)

M = [0.0524 +0.406]/115

pH = 0.004

Ph = -log[0.004]

Ph = 2.4