Question

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

(a) 0.00809 M HBr

pH = _____

(b) 0.260 g of HNO₃ in 36.0 L of solution

pH = _____

(c) 70.0 mL of 8.90 M HBr diluted to 1.80 L

pH = _____

(d) a mixture formed by adding 63.0 mL of 0.00678 M HBr to 84.0 mL of 0.00612 M HNO₃

pH = _____

Answer 1

(a) 0.00809 M HBr

HBr is a strong acid

so, [H⁺] = 0.00809 M

pH = -log [H⁺] = -log (0.00809) = 2.09

(b) (b) 0.260 g of HNO3 in 36.0 L of solution

molar mass of HNO3 = 63.01 g/mol

so, 0.260 g of HNO3 = 0.00413 mole

Volume = 36 L

[HNO3] = (0.00413 mole) / (36 L) = 1.15 x 10^-4 M.

(c) 70.0 mL of 8.90 M HBr diluted to 1.80 L

Moles = molarity x volume

Moles of HBr = 0.07 L x 8.9 M = 0.623 moles

Now, 0.623 moles of HBr diluted to 1.80 L

[HBr] = (0.623 moles) / (1.80 L) = 0.35 M

pH = -log (0.35) = 0.46

(d) a mixture formed by adding 63.0 mL of 0.00678 M HBr to 84.0 mL of 0.00612 M HNO3

Moles of H⁺ in HBr = (0.063 x 0.00678) = 0.000427 moles

Moles of H⁺ in HNO3 = (0.084 x 0.00612) = 0.000514 moles

Total moles of H⁺ = 0.000427 moles +  0.000514 moles = 0.000941 moles

Total volume = 63.0 mL + 84.0 mL = 147 mL = 0.147 L

[H⁺] = 0.000941 moles / 0.147 L = 0.0064 M

pH = -log(0.0064) = 2.19.