What is the pH of the solution formed by mixing 100 mL of 0.5 mol/L ammonia with 50 mL of 1 M hydrochloric acid?

For ammonia. #"p"K_text(a) = 9.25#.

1 Answer
Nov 27, 2017

Answer:

pH = 4.9

Explanation:

Step 1. Figure out what is happening

The #"p"K_text(a)# values tell you that #"HCl"# is a strong acid, that #"NH"_4^"+"# is a weak acid, and that #"NH"_3# is a weak base.

Thus, when you mix the strong acid #"HCl"# with the weak base #"NH"_3#, the reaction will go to completion.

#"Initial moles of NH"_3 = 0.100 color(red)(cancel(color(black)("L NH"_3))) × "0.5 mol NH"_3/(1 color(red)(cancel(color(black)("L NH"_3)))) = "0.050 mol NH"_3#

#"Initial moles of HCl" = 0.050 color(red)(cancel(color(black)("L HCl"))) × "1 mol HCl"/(1 color(red)(cancel(color(black)("L HCl")))) = "0.050 mol HCl"#

#color(white)(mmmmml)"NH"_3 +color(white)(l) "HCl" → "NH"_4"Cl"#
#"I/mol":color(white)(mll)0.050color(white)(ml)0.050color(white)(mmml)0#
#"C/mol":color(white)(m)"-0.050"color(white)(m)"-0.050"color(white)(m)"+0.050#
#"E/mol":color(white)(mmll)0color(white)(mmml)0color(white)(mmml)0.050#

The ammonia and hydrochloric acid react completely.

You now have 0.050 mol of #"NH"_4"Cl"# in 150 mL of solution.

Step 2. Calculate the pH of the solution

#"Initial concentration of NH"_4^"+" = "0.050 mol"/"0.150 L" = "0.33 mol/L"#

Now, we can use an ICE table.

#color(white)(mmmmmml)"NH"_4^"+" + "H"_2"O" ⇌ "NH"_3 + "H"_3"O"^"+"#
#"I/mol·L"^"-1":color(white)(ml)0.33 color(white)(mmmmmml)0color(white)(mmml)0#
#"C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mml)"+"x#
#"E/mol·L"^"-1":color(white)(m)"0.33-"xcolor(white)(mmmmml)xcolor(white)(mmml)x#

#K_text(a) = (["NH"_3]["H"_3"O"^"+"])/(["NH"_4^"+"]) = x^2/(0.33-x)#

#"p"K_text(a) = 9.25#

#K_text(a) = 10^("-p"K_text(a)) = 10^"-9.25" = 5.62 × 10^"-10"#

#x^2/(0.33-x) = 5.62 × 10^"-10"#

#0.33/(5.62 × 10^"-10") = 5.9 ×10^8#, so #x≪ 0.33#

Then,

#x^2/0.33 = 5.62 × 10^"-10"#

#x^2 = 0.33 × 5.62 × 10^"-10" = 1.9 × 10^"-10"#

#x = 1.4 × 10^"-5"#

#["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = 1.4 × 10^"-5"color(white)(l) "mol/L"#

#"pH" = "-log"["H"_3"O"^"+"] = -log(1.4 × 10^"-5") =4.9#