What is the pH at the equivalence point when 35.0 mL of 0.20 M ammonia is titrated by 0.12M hydrochloric acid? Kb for ammonia is 1.8 x 10-5.

Even without doing any calculations, you can say that the pH of the solution at equivalence point will be smaller than #7#.

In order to reach the equivalence point, you must add just enough acid to neutralize the base. #color(red)("(*)")#

In the case of a weak base being titrated with a strong acid, the neutralization reaction will produce the conjugate acid of the weak base, which will then react with water to produce *hydronium ions, #"H"_3"O"^(+)#, and reform some of the weak base.

The formation of the hydronium ions will cause the pH of the solution to be smaller than #7# at equivalence point, i.e. the resulting solution will be acidic.

So, ammonia reacts with hydrochloric acid according to the balanced chemical equation

#"NH"_text(3(aq]) + "HCl"_text((aq]) -> "NH"_text(4(aq])^(+) + "Cl"_text((aq])^(-)#

Notice that you have #1:1# mole ratios between the species that take part in the reaction. This tells you that reaction will

• consume one mole of hydrochloric acid for every one mole of ammonia
• produce one mole of ammonium ions for every one mole of ammonia consumed

Use the ammonia solution's molarity and volume to determine how many moles of ammonia you have in the initial solution

#color(blue)(c = n/V implies n = c * V)#

#n_(NH_3) = "0.20 M" * 35.0 * 10^(-3)"L" = "0.0070 moles NH"_3#

Now, according to #color(red)("(*)")# and to the balanced chemical equation, the equivalence point requires equal numbers of moles of ammonia and hydrochloric acid.

This means that you must add

#color(blue)(c = n/V implies V = n/c)#

#V_(HCl) = (0.0070 color(red)(cancel(color(black)("moles"))))/(0.12color(red)(cancel(color(black)("moles")))/"L") = "0.05833 L" = "58.3 mL"#

of hydrochloric acid to the ammonia solution.

The total volume of the solution will thus be

#V_"total" = V_(NH_3) + V_(HCl)#

#V_"total" = "35.0 mL" + "58.3 mL" = "93.3 mL"#

So, if #0.0070# moles of ammonia and #0.0070# moles of hydrochloric acid were consumed, it follows that the reaction produced #0.0070# moles of ammonium ions.

The concentration of the ammonium ions will be

#["NH"_4^(+)] = "0.0070 moles"/(93.3 * 10^(-3)"L") = "0.07503 M"#

Now, use an ICE table to help you find the equilibrium concentration of hydronium ions

#" " "NH"_text(4(aq])^(+) + "H"_2"O"_text((l]) " " rightleftharpoons " " "NH"_text(3(aq]) " "+" " "H"_3"O"_text((aq])^(+)#

#color(purple)("I") " "color(white)(x)0.07503" " " " " " " " " " " " " " " " "0" " " " " " " " " " " " "0#
#color(purple)("C") " "color(white)(x)(-x)" " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x)#
#color(purple)("E") " "0.07503-x" " " " " " " " " " " " " "x" " " " " " " " " " " " "x#

Now, notice that the problem provides you with the base dissociation constant, #K_b#. In order to find the acid dissociation constant, #K_a#, for the ammonium ion, use the equation

#color(blue)( K_a * K_b = K_W)" "#, where

#K_W# - the self-ionization constant of water, equal to #10^(-14)# at room temperature

This means that you have

#K_a = K_W/K_b = 10^(-14)/(1.8 * 10^(-5)) = 5.6 * 10^(-10)#

By definition, the acid dissociation constant is equal to

#K_a = ( ["NH"_3] * ["H"_3"O"^(+)])/(["NH"_4^(+)]) = (x * x)/(0.07503 -x)#

Since #K_a# has such a small value, you can say that

#0.07503 - x ~~ 0.07503#

This means that you have

#K_a = x^2/0.07503 = 5.6 * 10^(-10)#

Therefore,

#x = sqrt( 0.07503 * 5.6 * 10^(-10)) = 6.5 * 10^(-6)#

Since #x# represents the equilibrium concentration of the hydronium ions, it follows that you will have

#["H"_3"O"^(+)] = 6.5 * 10^(-6)"M"#

The pH of the solution will thus be

#"pH" = - log( ["H"_3"O"^(+)])#

#"pH" = - log( 6.5 * 10^(-6)) = color(green)(5.19)#

As predicted, the pH of the resulting solution will be smaller than #7# at equivalence point.

SIDE NOTE As practice, you can try using water's concentration of hydronium ions, #10^(-7)"M"#, in the ICE table to see if this changes the value of the pH.

From what I can tell, it will have little impact, if any, on the final pH of the solution.