For a solution containing initial concentrations of #"0.0500 M"# #"HNO"_2# and #"0.0100 M HClO"#, what is the #"pH"#?
1 Answer
Explanation:
!! LONG ANSWER !!
The idea here is that you're mixing two weak acids, so right from the start you know that you're dealing with equilibrium reactions.
Before moving forward, make sure that you have the acid dissociation constants,
So, you know that
#K_a = 4.0 * 10^(-4) -># for nitrous acid
#K_a = 3.0 * 10^(-8) -># for hypochlorous acid
Notice that there is quite a significant difference between the orders of magnitude of these two acid dissociation constants in favor of a greater quantity of the nitrous acid dissociating.
Two equilibrium reactions will be established in aqueous solution
#"HNO"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NO"_ (2(aq))^(-) + "H"_ 3"O"_((aq))^(+)#
#"HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "ClO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)#
Now, the magnitude of the acid dissociation constant essentially tells you how far to the left will each ionization equilibrium lie.
More specifically, the smaller the value of
You can compare the two acid dissociation constants relative to each other.
Since hypochlorous acid has the smaller
Now, more molecules of nitrous acid will ionize to form nitrite anions,
This tells you that more hydronium cations will come from the ionization of the nitrous acid than from the ionization of the hypochlorous acid.
At this point, you can think about the hydronium cation as being a common ion. Its presence in solution will affect the position of the equilibrium for the ionization of the hypochlorous acid
More specifically, the presence of the hydronium cations that are coming from the ionization of the nitrous acid will push the ionization equilibrium of the hypochlorous acid even further to the left.
As a result, even fewer hydronium cations will be delivered to the solution by the hypochlorous acid.
This means that you can predict that the pH of the solution will actually be, for all intended purposes, equal to the pH of a solution that contains solely the nitrous acid.
So, use an ICE table to find the equilibrium concentration of hydronium ions coming from the ionization of the nitrous acid
#" ""HNO"_ (2(aq)) + "H"_ 2"O"_ ((l)) " "rightleftharpoons" " "NO"_ (2(aq))^(-) + "H"_ 3"O"_((aq))^(+)#
By definition, the acid dissociation constant will be
#K_a = (["NO"_2^(-)] * ["H"_3"O"^(+)])/(["HNO"_2])#
In your case, you will have
#4.0 * 10^(-4) = (x * x)/(0.0500 - x) = x^2/(0.0500 - x)#
Rearrange to form a quadratic equation
#x^2 + 4.0 * 10^(-4)* x - 0.2 * 10^(-4) = 0#
This will produce two solutions, one positive and one negative. Since
You will have
#x = 0.00428#
This means that the equilibrium concentration of hydronium ions will be
#["H"_3"O"^(+)] = x = "0.00428 M"#
Now, use this concentration of hydronium ions as a starting concentration in the ionization equilibrium of hypochlorous acid. You will have
#" ""HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) " "rightleftharpoons" " "ClO"_ ((aq))^(-) +" " "H"_ 3"O"_ ((aq))^(+)#
This time, you will have
#K_a = (["ClO"^(-)] * ["H"_3"O"^(+)])/(["HClO"])#
which in your case is
#3.0 * 10^(-8) = (x * (0.00428 + x ))/(0.0100 -x)#
This time, the value of
#0.00428 + x ~~ 0.00428" "# and#" "0.0100 - x ~~ 0.0100#
This will get you
#3.0 * 10^(-8) = x * 0.00428/0.00100 => x = 7.01 * 10^(-8)# (if you had not made the small
#K_a# approximation, you would have gotten about#7.0092 * 10^(-8)# , which you could see is close enough to the unrounded approximated answer of#7.0093 * 10^(-8)# .)
The ionization of the hypochlorous acid will thus produce
#["H"_3"O"^(+)] = 7.01 * 10^(-8)"M"#
As you can see, the prediction that the pH of the solution will be approximately equal to that of a solution containing solely nitrous acid turns out to be a valid one, since
#["H"_ 3"O"^(+)]_"TOTAL" = "0.00428 M" + 7.01 * 10^(-8)"M" ~~ "0.00428 M"#
Since you know that
#color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))#
you can say that the pH of the solution will be
#"pH" = - log(0.00428) = color(green)(|bar(ul(color(white)(a/a)2.37color(white)(a/a)|)))#