Question

Calculate the pH of each of the following strong acid solutions. (a) 0.00227 M HIO 4 pH = 2.64 (b) 0.724 g of HBrO4 in 49.0 L of solution pH = lý (c) 13.0 mL of 7.40 M HIO4 diluted to 2.60 L pH = 3 (d) a mixture formed by adding 60.0 mL of 0.00580 M HIO4 to 81.0 mL of 0.00341 M HBrO pH =

Answer 1

b) First calculate [HBrO₄]

Molar mass of HBrO₄ = 1.0079 + 79.91 + ( 4 $\times$ 16.00) = 144.92 g / mol

Mass of HBrO₄ = 0.724 g

We have, No. of moles = Mass / Molar mass

$\therefore$ No. of moles of HBrO₄ = 0.724 g / 144.92 g /mol = 0.004996 mol

volume of solution = 49.0 L

[ HBrO₄ ] = No. of moles of  HBrO₄ / volume in L = 0.004996 mol / 49.0 L = 0.000102 M

We have, pH = - log [H₃O⁺] = - log 0.000102 = 3.99

ANSWER : pH = 3.99

C) We can use dilution formula to calculate molarity of diluted solution of HIO₄

We have formula, M stock $\times$ V stock = M dilute $\times$ V dilute

$\therefore$ M dilute = M stock $\times$ V stock / V dilute

M dilute = 7.40 M $\times$ 13.0 ml / 2600 ml = 0.037 M

[ HIO₄ ] dilute = 0.037 M

We have, pH = - log [H₃O⁺] = - log 0.037 =1.43

ANSWER : pH =1.43

D) mmol of H ⁺ from HIO₄ = concentration $\times$ volume = 0.00580 $\times$ 60.0 = 0.348 mmol

mmol of H ⁺ from HBrO₄ = 0.00341 $\times$ 81.0 = 0.2762 mmol

Total mmol of H ⁺ in the acid mixture = 0.348 + 0.2762 = 0.6242 mmol

Total volume of acid mixture = 60.0 + 81.0 = 141.0 ml

$\therefore$ [H ⁺ ] = No .of moles of H ⁺ / volume of solution in L

[H ⁺ ] = [ 0.6242 / 1000] / [ 141.0 / 1000]= 0.00443 M

We have, pH = - log [H ⁺ ] = - log 0.00443 =2.36

ANSWER : pH =2.36