chemistry calculate pH

Question

Calculate the pH when 90.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH₃NH₂ (Kb = 4.4 × 10⁻⁴)

Answer 1

Write the balanced equation for the reaction:

HBr + CH₃NH₂ <==> CH₃NH₃⁺Br^-

HBr is a strong acid. CH₃NH₂ is a weak base. CH₃NH₃⁺ is the conjugate acid of CH₃NH₂.

Since we have the amounts of both reactants, we must find which reactant is present in limiting supply.

moles HBr initially present = 90.0 ml x 1 L / 1000 ml x 0.200 mol / L = 0.018 mols

moles CH₃NH₂ initially present = 30.0 ml x 1 L / 1000 ml x 0.400 mol / L = 0.012 mols

The mole ratio in the balanced equation for HBr : CH₃NH₂ is 1 : 1 so we see that HBr + CH₃NH₂ is limiting.

Set up an ICE table:

HBr + CH₃NH₂ <==> CH₃NH₃⁺Br^-

0.018....0.012...............0.................Initial

-0.012.....-0.012.............+0.012..........Change

0.006.......0...................0.012..........Equilibrium

You can see that at equilibrium, we have only HBr (0.006 moles) and CH₃NH₃⁺ (0.012 moles). The total volume is 90 ml + 30 ml = 120 ml = 0.120 L. Final concentrations are:

[HBr] = 0.004 mol / 0.120 L = 0.05 M

[CH₃NH₃⁺] = 0.012 mol / 0.120 L = 0.1 M

To find the pH of this solution, we need to find the total [H⁺].

[H⁺] from HBr = 0.05 M (strong acid dissociates completely)

[H⁺] from CH₃NH₃⁺ is calculated from hydrolysis of CH₃NH₃⁺, but since Kb is 10^-4 the Ka for the conjugate acid will be on the order of 10^-11 which is such a weak acid, we need not consider it in the calculation of pH.

Thus, pH will be determined by the remaining HBr.

pH = -log [H+] = -log 0.05 M

pH = 1.3