For a solution containing initial concentrations of $0.0500 M$ $"0.0500 M"$ ${HNO}_{2}$ $"HNO"_2$ and $0.0100 M HClO$ $"0.0100 M HClO"$ , what is the $pH$ $"pH"$ ?

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For a solution containing initial concentrations of $0.0500 M$ $"0.0500 M"$ ${HNO}_{2}$ $"HNO"_2$ and $0.0100 M HClO$ $"0.0100 M HClO"$ , what is the $pH$ $"pH"$ ?

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Answer 1 · 2016-04-05T18:47:51

$pH = 2.37$ $"pH" = 2.37$

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, $K_{a}$ $K_a$ , for nitrous acid, ${HNO}_{2}$ $"HNO"_2$ , and hypochlorous acid, $HClO$ $"HClO"$ , which you can find listed here

http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

So, you know that

$K_{a} = 4.0 \cdot 10^{- 4} \to$ $K_a = 4.0 * 10^(-4) ->$ for nitrous acid

$K_{a} = 3.0 \cdot 10^{- 8} \to$ $K_a = 3.0 * 10^(-8) ->$ 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

${HNO}_{2 (a q)} + H_{2} O_{(l)} ⇌ {NO}_{2 (a q)}^{-} + H_{3} O_{(a q)}^{+}$ $"HNO"_ (2(aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "NO"_ (2(aq))^(-) + "H"_ 3"O"_((aq))^(+)$

${HClO}_{(a q)} + H_{2} O_{(l)} ⇌ {ClO}_{(a q)}^{-} + H_{3} O_{(a q)}^{+}$ $"HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "ClO"_ ((aq))^(-) + "H"_ 3"O"_ ((aq))^(+)$

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 $K_{a}$ $K_a$ , the more the equilibrium will lie to the left, i.e. the more acid molecules will not ionize.

You can compare the two acid dissociation constants relative to each other.

Since hypochlorous acid has the smaller $K_{a}$ $K_a$ , its ionization equilibrium will lie further to the left than that of the nitrous acid. This means that fewer hypochlorous acid molecules will actually ionize to form hypochlorite anions, ${ClO}^{-}$ $"ClO"^(-)$ , and hydronium cations, $H_{3} O^{+}$ $"H"_3"O"^(+)$ .

Now, more molecules of nitrous acid will ionize to form nitrite anions, ${NO}_{2}^{-}$ $"NO"_2^(-)$ , and hydronium cations.

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 $\to$ $->$ think Le Chatelier's Principle here.

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

${HNO}_{2 (a q)} + H_{2} O_{(l)} ⇌ {NO}_{2 (a q)}^{-} + H_{3} O_{(a q)}^{+}$ $" ""HNO"_ (2(aq)) + "H"_ 2"O"_ ((l)) " "rightleftharpoons" " "NO"_ (2(aq))^(-) + "H"_ 3"O"_((aq))^(+)$

$I a a a a a 0.0500 a a a a a a a a a a a a a a a a a a a 0 a a a a a a a a 0$ $color(purple)("I")color(white)(aaaaacolor(black)(0.0500)aaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaacolor(black)(0)$
$C a a a a a (- x) a a a a a a a a a a a a a a a a a (+ x) a a a a (+ x)$ $color(purple)("C")color(white)(aaaaacolor(black)((-x))aaaaaaaaaaaaaaaaacolor(black)((+x))aaaacolor(black)((+x))$
$E a a a 0.0500 - x a a a a a a a a a a a a a a a a a x a a a a a a a a x$ $color(purple)("E")color(white)(aaacolor(black)(0.0500-x)aaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaacolor(black)(x)$

By definition, the acid dissociation constant will be

$K_{a} = \frac{[{NO}_{2}^{-}] \cdot [H_{3} O^{+}]}{[{HNO}_{2}]}$ $K_a = (["NO"_2^(-)] * ["H"_3"O"^(+)])/(["HNO"_2])$

In your case, you will have

$4.0 \cdot 10^{- 4} = \frac{x \cdot x}{0.0500 - x} = \frac{x^{2}}{0.0500 - x}$ $4.0 * 10^(-4) = (x * x)/(0.0500 - x) = x^2/(0.0500 - x)$

Rearrange to form a quadratic equation

$x^{2} + 4.0 \cdot 10^{- 4} \cdot x - 0.2 \cdot 10^{- 4} = 0$ $x^2 + 4.0 * 10^(-4)* x - 0.2 * 10^(-4) = 0$

This will produce two solutions, one positive and one negative. Since $x$ $x$ represents concentration, the negative one does not carry any physical significance here.

You will have

$x = 0.00428$ $x = 0.00428$

This means that the equilibrium concentration of hydronium ions will be

$[H_{3} O^{+}] = x = 0.00428 M$ $["H"_3"O"^(+)] = x = "0.00428 M"$

Now, use this concentration of hydronium ions as a starting concentration in the ionization equilibrium of hypochlorous acid. You will have

${HClO}_{(a q)} + H_{2} O_{(l)} ⇌ {ClO}_{(a q)}^{-} + H_{3} O_{(a q)}^{+}$ $" ""HClO"_ ((aq)) + "H"_ 2"O"_ ((l)) " "rightleftharpoons" " "ClO"_ ((aq))^(-) +" " "H"_ 3"O"_ ((aq))^(+)$

$I a a a a 0.0100 a a a a a a a a a a a a a a a a a a a 0 a a a a a a a a a 0.00428$ $color(purple)("I")color(white)(aaaacolor(black)(0.0100)aaaaaaaaaaaaaaaaaaacolor(black)(0)aaaaaaaaacolor(black)(0.00428)$
$C a a a (- x) a a a a a a a a a a a a a a a a a a (+ x) a a a a a a a (+ x)$ $color(purple)("C")color(white)(aaacolor(black)((-x))aaaaaaaaaaaaaaaaaacolor(black)((+x))aaaaaaacolor(black)((+x))$
$E a a 0.0100 - x a a a a a a a a a a a a a a a a a x a a a a a a a a 0.00428 + x$ $color(purple)("E")color(white)(aacolor(black)(0.0100-x)aaaaaaaaaaaaaaaaacolor(black)(x)aaaaaaaacolor(black)(0.00428 + x)$

This time, you will have

$K_{a} = \frac{[{ClO}^{-}] \cdot [H_{3} O^{+}]}{[HClO]}$ $K_a = (["ClO"^(-)] * ["H"_3"O"^(+)])/(["HClO"])$

which in your case is

$3.0 \cdot 10^{- 8} = \frac{x \cdot (0.00428 + x)}{0.0100 - x}$ $3.0 * 10^(-8) = (x * (0.00428 + x ))/(0.0100 -x)$

This time, the value of $K_{a}$ $K_a$ is so small that you can safely use the approximations

$0.00428 + x \approx 0.00428$ $0.00428 + x ~~ 0.00428" "$ and $0.0100 - x \approx 0.0100$ $" "0.0100 - x ~~ 0.0100$

This will get you

$3.0 \cdot 10^{- 8} = x \cdot \frac{0.00428}{0.00100} \Rightarrow x = 7.01 \cdot 10^{- 8}$ $3.0 * 10^(-8) = x * 0.00428/0.00100 => x = 7.01 * 10^(-8)$

(if you had not made the small $K_{a}$ $K_a$ approximation, you would have gotten about $7.0092 \cdot 10^{- 8}$ $7.0092 * 10^(-8)$ , which you could see is close enough to the unrounded approximated answer of $7.0093 \cdot 10^{- 8}$ $7.0093 * 10^(-8)$ .)

The ionization of the hypochlorous acid will thus produce

$[H_{3} O^{+}] = 7.01 \cdot 10^{- 8} M$ $["H"_3"O"^(+)] = 7.01 * 10^(-8)"M"$

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

${[H_{3} O^{+}]}_{TOTAL}^{+} = 0.00428 M + 7.01 \cdot 10^{- 8} M \approx 0.00428 M$ $["H"_ 3"O"^(+)]_"TOTAL" = "0.00428 M" + 7.01 * 10^(-8)"M" ~~ "0.00428 M"$

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

$"pH" = - log(0.00428) = color(green)(|bar(ul(color(white)(a/a)2.37color(white)(a/a)|)))$

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For a solution containing initial concentrations of $0.0500 M$ $"0.0500 M"$ ${HNO}_{2}$ $"HNO"_2$ and $0.0100 M HClO$ $"0.0100 M HClO"$ , what is the $pH$ $"pH"$ ?

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