Question #bdf57
1 Answer
Explanation:
The idea here is that the ammonium cation,
Ammonium chloride,
#"NH"_ 4"Cl"_ ((aq)) -> "NH"_ (4(aq))^(+) + "Cl"_((aq))^(-)#
Since the salt dissociates in a
#["NH"_4^(+)] = ["NH"_4"Cl" ] = "0.1 mol dm"^(-3)#
Now, in order to be able to calculate the pH of the solution, you need to know the acid dissociation constant,
#K_a = 5.6 * 10^(-10)#
Once in aqueous solution, the ammonium cations will react with water to form ammonia,
Now, because the acid dissociation constant is so small, let us at first include the dissociation of water, given by
We now use an ICE table to help you find the equilibrium concentration of the hydronium cations.
#" " "NH"_ (4(aq))^(+) + "H"_ 2"O"_ ((l)) " "rightleftharpoons" " "NH"_ (3(aq)) + "H"_ 3"O"_((aq))^(+)#
By definition, the acid dissociation constant will be
#K_a = (["H"_3"O"^(+)] * ["NH"_3])/(["NH"_4^(+)])#
In your case, this will be equivalent to
#K_a = (x * (1 * 10^(-7) + x))/(0.1 -x ) = 5.6 * 10^(-10)#
Once again, because
#0.1 - x ~~ 0.1#
This will get you
#5.6 * 10^(-10) = ( x * (x+y))/0.1#
#color(purple){x * (x+y)=5.6 * 10^{-11}}#
Now we need another equation to get
#["H"_3"O"^+]["O""H"^-]=color(brown){(x+y) * y=1.0 * 10^{-14}}#
We divide the brown equation by the purple one and see that the factor
#y=0.000179x#
And out this back into the purple equation for the ammonium ion hydrolysis:
#color(purple){x * (x+0.00018x)=1.000179x^2=5.6 * 10^{-11}}#
Then take the square root. Remember that
#x = \sqrt{{5.6 * 10^(-11)}/1.000179}=7.48 * 10^{-6} M#
Observe that the
The total concentration of hydronium cations present in solution at equilibrium will be
#["H"_3"O"^(+)] = overbrace(7.48 * 10^(-6)"M")^(color(purple)("x, coming from NH"_4^(+))) + overbrace(1.34 * 10^(-9)"M")^(color(brown)("y, coming from the self-ionization of water"))#
#["H"_3"O"^(+)] = 7.48 * 10^(-6)"M"#
Here
#"M" = "mol dm"^(-3)#
As you know, pH is calculated using
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
In your case, you will have
#"pH" = - log(7.47 * 10^(-6)) = color(green)(|bar(ul(color(white)(a/a)5.13color(white)(a/a)|)))#
SIDE NOTE Even with this small dissociation constant, there is so little water dissociation that the concentration of hydronium ion,