For a): Ammonium chloride, #NH_4Cl# dissolves in solution to form ammonium ions #NH_4^(+)# which act as a weak acid by protonating water to form ammonia, #NH_3(aq)# and hydronium ions #H_3O^(+)(aq)#: #NH_4^(+)(aq)+H_2O(l) -> NH_3(aq)+H_3O^(+)(aq)# As we know the #K_b# for ammonia, we can find the #K_a# for the ammonium ion. For a given acid/base pair: #K_a times K_b=1.0 times 10^-14# assuming standard conditions. So, #K_a(NH_4^(+))=(1.0 times 10^-14)/(1.8 times 10^-5)=5.56 times 10^-10# Plug in the concentration and the #K_a# value into the expression: #K_a=([H_3O^(+)] times (NH_3])/([NH_4^(+)])# #5.56 times 10^-10~~([H_3O^(+)] times [NH_3])/([0.1])# #5.56 times 10^-11=[H_3O^(+)]^2# (as we can assume that one molecule hydronium must form for every one of ammonia that forms. Also, #K_a# is small, so #x ≪ 0.1#.) #[H_3O^(+)]=7.45 times 10^-6# #pH=-log[H_3O^(+)]# #pH=-log(7.45 times 10^-6)# #pH approx 5.13# For b): (i) Determine the species present after mixing. The equation for the reaction is #color(white)(mmmmm)"OH"^"-" + "NH"_4^"+" -> "NH"_3 + "H"_2"O"# #"Moles of OH"^"-" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol"# #"Moles of NH"_4^"+" = "0.100 L" × "0.1 mol"/"1 L" = "0.010 mol"# So, we will have 200 mL of an aqueous solution containing 0.010 mol of ammonia, and the pH should be higher than 7. (ii) Calculate the pH of the solution #["NH"_3] = "0.010 mol"/"0.200 L" = "0.050 mol/L"# The chemical equation for the equilibrium is #"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"# Let's re-write this as #"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"# We can use an ICE table to do the calculation. #color(white)(mmmmmmmll)"B" + "H"_2"O" ⇌ "BH"^"+" + "OH"^"-"# #K_text(b) = (["BH"^"+"]["OH"^"-"])/(["B"]) = x^2/("0.050-"x) = 1.8 × 10^"-5"# Check for negligibility: #0.050/(1.8 × 10^"-5") = 3 × 10^3 > 400#. ∴ #x ≪ 0.050# #x^2/0.050 = 1.8 × 10^"-5"# #x^2 = 0.050 × 1.8 × 10^"-5" = 9.0 × 10^"-7"# #x = 9.5 × 10^"-4"# #["OH"^"-"] = 9.5 × 10^"-4" color(white)(l)"mol/L"# #"pOH" = -log(9.5 × 10^"-4") = 3.0# #"pH = 14.00 - pOH = 14.00 - 3.0" = 11.0#
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