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QUESTION

# What is the pH of 0.1 M ammonia solution if Ka for the ammonium ion is 5.62 x 10-10 and Kw = 10-14?

p"H" = 11.13

The idea here is to use the ammonium ion's acid dissociation constant, K_a, to find the value of ammonia's base dissociation constant, K_b.

The relationship between the base dissocaition constant, the acid dissociation constant, and water's self-ionization constant, K_W, is given by the equation

K_W = K_a xx K_b

In your case, the base dissociation constant of ammonia will be

K_b = K_W/K_a = 10^(-14)/(5.62 * 10^(-10)) = 1.78 * 10^(-5)

To determine the of the solution, you first need to know the p"OH" of the solution, which in turn requires the cocnentration of hydroxide ions, "OH"^(-).

Ammonia will react with water to form ammonium and hydroxide ions. Use an ICE table to determine the equilibrium concentration of hydroxide ions

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

color(purple)("I")" " " " 0.1" " " " " " " " " " " " " " " " " " " "0" " " " " " " " " " " "0 color(purple)("C")" " color(white)(x)-x" " " " " " " " " " " " " " " " " "(+x)" " " " " " " " "(+x) color(purple)("E")" " color(white)(x)0.1-x" " " " " " " " " " " " " " " "color(white)(xx)x" " " " " " " " " "color(white)(x)x

By definition, the base dissociation constant will be

K_b = (["NH"_4^(+)] * ["OH"^(-)])/(["NH"_3]) = (x * x)/(0.1 - x)

Because the value of K_b is so small, you can say that

(0.1 - x) ~~ 0.1

This means that you have

K_b = x^2/0.1 = 1.78 * 10^(-5)

x = sqrt(0.1 * 1.78 * 10^(-5)) = 0.001334

This means that you have

["OH"^(-)] = x = "0.001334 M"

The p"OH" of the solution will be

p"OH" = -log(["OH"^(-)])

p"OH" = -log(0.001334) = 2.87

The pH of the solution will thus be

p"H" = 14 - p"OH" = 14 - 2.87 = color(green)(11.13)

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