Question

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



(a) 0.00851 M HCl

pH =  



(b) 0.714 g of HNO₃ in 18.0 L of solution

pH =  



(c) 62.0 mL of 4.90 M HCl diluted to 3.00 L

pH =  



(d) a mixture formed by adding 55.0 mL of 0.00326 M HCl to 46.0 mL of 0.00896 M HNO₃

pH =  

Answer 1

(a) 0.00851 M HCl

[H⁺] = 0.00851 M

pH = -log[H⁺] = 2.07

(b) 0.714 g of HNO₃ in 18.0 L of solution

Molar mass of HNO₃ = 63.01 g/mol

0.714 g HNO₃ = 0.011 mole HNO₃

Molarity of HNO₃ solution

= 0.011/18 = 6.3 * 10^-4 M

[H⁺] =6.3 * 10^-4 M

pH = -log[H⁺] = 3.2

(c) 62.0 mL of 4.90 M HCl diluted to 3.00 L

Apply equation, V₁S₁ = V₂S₂ to determine the strength of HCl in the resulting solution.

S₂ = strength of HCl in the resulting solution

So, 62.0 * 4.90 = 30000 * S₂

S₂ = 0.10 M

[H⁺] =0.10 M

pH = -log[H⁺] = 1

(d) a mixture formed by adding 55.0 mL of 0.00326 M HCl to 46.0 mL of 0.00896 M HNO₃

Both are strong acids. Both of them will dissociate completely in the solution.

Total volume = 55 + 46 = 101 mL

Strength of HCl in the resulting solution = 55*0.00326/ 101 = 1.77 * 10^-3 M

Strength of HNO₃ in the resulting solution = 46*0.00896/ 101 = 4.08 * 10^-3 M

[H⁺] =1.77 * 10^-3 M + 4.08 * 10^-3 M = 5.85 * 10^-3 M

pH = -log[H⁺] = 2.23