Question

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

(a) 0.00593 M HIO₄

pH = ____

(b) 0.670 g of HCl in 50.0 L of solution

pH = ____

(c) 75.0 mL of 1.00 M HIO₄ diluted to 2.30 L

pH = ____

(d) a mixture formed by adding 83.0 mL of 0.00358 M HIO₄ to 30.0 mL of 0.000920 M HCl

pH = ____

Answer 1

(a) 0.00593 M HIO₄

HIO4 is a strong acid so it only form ion in water disolution

[H+] = 0.00593 M

pH = -log[H+]

pH = 2.227

(b)  0.670 g of HCl in 50.0 L of solution

concentration of HCl = (no. of moles / volume in liter)

moles of HCl = (0.67g / 36.5g/mol) = 0.01836 moles

molarity of HCl = 0.01836 moles / 50.0 L = 0.000367 M

HC is strong acid so In aqueous solution fully dissociates to H+ and Cl−

[H+] = 0.000367 M

pH = -log[H+] = 3.435

(C) 75.0 mL of 1.00 M HIO4 diluted to 2.30 L

number of moles of HIO4 = (1.00 M x 0.075 L) = 0.075 moles

concentration after diluted = ( 0.075 moles / 2.30 L ) = 0.0326 M

pH = -log [H+] = -log[0.0326]

pH = 1.487

(d)  a mixture formed by adding 83.0 mL of 0.00358 M HIO4 to 30.0 mL of 0.000920 M HCl

both are strong acid so only concentration will resultant M1V1 + M2V2 = MV

M = [ (0.083L x  0.00358 M) +(0.03 L x 0.000920 M) ] / 0.113 L = 0.002874 M

pH = -log [0.002874] = 2.5415