Question

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



(a) 0.00438 M HClO₄

pH =



(b) 0.579 g of HCl in 21.0 L of solution

pH =



(c) 59.0 mL of 3.90 M HClO₄ diluted to 4.50 L

pH =



(d) a mixture formed by adding 69.0 mL of 0.00809 M HClO₄ to 78.0 mL of 0.00483 M HCl

pH =

Answer 1

pH = -log[H+]

(a)

HClO4 is strong acid so [HClO4] = [H+]

pH of 0.00438 M HClO4

pH = -log[0.00438]

pH = 2.36

(b)

Molarity = moles of solute / Liters of solution

Moles of HCl = 0.579 g HCl × ( 1 mol HCl/ 36.46 g HCl)

= 0.0159 mol HCl

Molarity of HCl = 0.0159 mol / 21.0 L

= 0.000757 M

pH = - log(0.000757)

= 3.12

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(C)

Dilution principle:

M1V1 = M2V2

M1 = stock solution molarity

V1 = stock Solution volume

M2 = dilute solution molarity

V2 = dilute solution volume

Concentration of HClO4

= 3.90 M × 0.059 L / 4.50 L

= 0.0511 M

pH = -log[0.0511]

= 1.29

(d)

Resultant solution concentration

= (M1V1 + M2V2) / (V1+V2)

= 0.00636 M

pH = -log(0.00636)

= 2.20

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