Question

A. Calculate the pH of an HCl solution in pure water that’s initially 5.0 x 10–8 M. Does the value you obtain appear to follow the trend you see in Table 2?

B. Calculate the pH of an aqueous solution that’s initially 5.0 x 10–8 M in HCl and 0.10 M in NaCl. Don’t forget that activities need to be used in the Kw expression but actual concentrations are called for when balancing positive and negative charges.

Answer 1

1) HCl(aq) --------> H⁺(aq) + Cl^-(aq)

[H⁺] = [Cl^-] = [HCl] = 5*10^-8 M

Also,

H₂O(l) <--------> H⁺(aq) + OH^-(aq)

[H⁺] = [OH^-] = 10^-7 M ; (as K_w = [H⁺]*[OH^-] = 10^-14 & [H⁺] = [OH^-] = 10^-7 M)

Total [H⁺] = 5*10^-8 + 1*10^-7 = 1.5*10^-7 M

Thus, pH = -log[H⁺] = 6.824

2)

1) HCl(aq) --------> H⁺(aq) + Cl^-(aq)

[H⁺] = [Cl^-] = [HCl] = 5*10^-8 M

Also,

NaCl(aq) --------> Na⁺(aq) + Cl^-(aq)

H₂O(l) <--------> H⁺(aq) + OH^-(aq)

[H⁺] = [OH^-] = 10^-7 M ; (as K_w = [H⁺]*[OH^-] = 10^-14 & [H⁺] = [OH^-] = 10^-7 M)

Total [H⁺] = 5*10^-8 + 1*10^-7 = 1.5*10^-7 M

Thus, pH = -log[H⁺] = 6.824