What is the pH of a solution of 10^(-8)10−8 M sodium hydroxide ?

2 Answers
Stefan V.
Aug 12, 2016

"pH" = 7.02pH=7.02

Explanation:

Right from the start, you should be able to predict that the pH of the solution will be higher than 77. This is the case because the solution contains a strong base.

You could look at the concentration of the base and say that the pH will only be slightly higher than 77, but it must come out to be higher than 77.

The key here is the self-ionization of water, which at room temperature has an equilibrium constant equal to

K_W = 10^(-14) ->KW=10−14→ water's ionization constant

In pure water, water undergoes self-ionization to form hydronium cations, "H"_3"O"^(+)H3O+ and hydroxide anions, "OH"^(-)OH−, as described by the following equilibrium reaction

2"H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "OH"_ ((aq))^(-)2H2O(l)⇌H3O+(aq)+OH−(aq)

By definition, the ionization constant is equal to

K_W = ["H"_3"O"^(+)] * ["OH"^(-)]KW=[H3O+]⋅[OH−]

If you take xx to be the equilibrium concentration of hydronium cations and hydroxide anions in pure water, you can say that you have

K_w = x * x = x^2Kw=x⋅x=x2

This gets you

x = sqrt(K_W) = sqrt(10^(-14)) = 10^(-7)x=√KW=√10−14=10−7

So, the self-ionization of water produces

["H"_3"O"^(+)] = 10^(-7)"M "[H3O+]=10−7M and " " ["OH"^(-)] = 10^(-7)"M" [OH−]=10−7M

at room temperature. Now, your solution contains sodium hydroxide, "NaOH"NaOH, a strong base that dissociates completely to produce hydroxide anions in a 1:11:1 mole ratio.

Therefore, you're adding

["OH"^(-)] = ["NaOH"] = 10^(-8)"M"[OH−]=[NaOH]=10−8M

to pure water. The key now is the fact that the self-ionization of water will still take place, but because the concentration of hydroxide anions has increased, the equilibrium will produce fewer moles of hydronium and hydroxide ions.

If you take yy to be the concentrations of hydronium and hydroxide ions produced by the self-ionization reaction, you can say that you have

" "2"H"_ 2"O"_ ((l)) rightleftharpoons" " "H"_ 3"O"_ ((aq))^(+) " "+" " " ""OH"_ ((aq))^(-) 2H2O(l)⇌ H3O+(aq) + OH−(aq)

color(purple)("I")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(0)aaaaaaaaaaaaacolor(black)(10^(-8))Iaaaaaa−aaaaaaaaa0aaaaaaaaaaaaa10−8
color(purple)("C")color(white)(aaaaaacolor(black)(-)aaaaaaacolor(black)((+y))aaaaaaaaacolor(black)((10^(-8) + y))Caaaaaa−aaaaaaa(+y)aaaaaaaaa(10−8+y)
color(purple)("E")color(white)(aaaaaacolor(black)(-)aaaaaaaaacolor(black)(y)aaaaaaaaaaacolor(black)(10^(-8) + y)Eaaaaaa−aaaaaaaaayaaaaaaaaaaa10−8+y

This time, the ionization constant will be equal to

K_W = y * (10^(-8) + y)KW=y⋅(10−8+y)

K_w = y^2 + 10^(-8) * yKw=y2+10−8⋅y

This will get you

y^2 + 10^(-8) * y - 10^(-14) = 0y2+10−8⋅y−10−14=0

This quadratic equation will produce two solutions, one positive and one negative. Since yy represents concentration, pick the positive one

y = 9.51 * 10^(-8)y=9.51⋅10−8

This means that the equilibrium concentration of hydronium cations will be

["H"_3"O"^(+)] = 9.51 * 10^(-8)"M"[H3O+]=9.51⋅10−8M

The pH of the solution is given by

color(blue)(|bar(ul(color(white)(a/a)"pH" = - log(["H"_3"O"^(+)])color(white)(a/a)|)))

Plug in your value to find

"pH" = - log(9.51 * 10^(-8)) = color(green)(|bar(ul(color(white)(a/a)color(black)(7.02)color(white)(a/a)|)))

As you can see, the pH of the solution is consistent with the fact that your solution contains a strong base, albeit in a very small concentration.

Michael
Aug 14, 2016

sf(pH=7.02)

Explanation:

This is a tiny concentration for NaOH so I would expect the pH to be close to 7 or slightly greater.

Because the concentration is so small, we must take into account the ions that are produced from the auto - ionisation of water:

sf(H_2OrightleftharpoonsH^++OH^-)

For which sf(K_w=[H^+][OH^-]=10^(-14)color(white)(x)"mol"^2."l"^(-2)) at sf(25^@C).

This tells us that, in pure water, the hydrogen and hydroxide ion concentrations are sf(10^(-7)color(white)(x)"mol/l") respectively.

To get the total hydroxide concentration you might think you simply add sf(10^(-7)) and sf(10^(-8)). This is not the case as these will not be equilibrium concentrations.

We need to apply Le Chatelier's Principle.

If we do a thought experiment we can imagine some pure water to which a tiny amount of sodium hydroxide is added such that the concentration of NaOH is sf(10^-8color(white)(x)"mol/l".

We have now disturbed a system at equilibrium by adding extra sf(OH^-) ions. The system will act to oppose that change by reducing the number of sf(OH^-) ions and shifting to the left.

The reaction quotient sf(Q) is given by sf([H^+][OH^-]). This is now greater than sf(K_w) so the system responds by shifting the position of equilibrium to the left such that sf(Q=K_w).

We can set up an ICE table using equilibrium concentrations to show this:

" "sf(H_2O" "rightleftharpoons" "H^+" "+" "OH^-)

sf(color(red)(I)color(white)(xxxxxxxxx)" "10^-7" "(10^(-7)+10^(-8))

sf(color(red)(C)color(white)(xxxxxxx)" "-xcolor(white)(xxx)" "-x)

sf(color(red)(E)color(white)(xxxxxxxx)" "(10^(-7)-x)" "(1.1xx10^(-7)-x))

This gives us:

sf((10^-7-x)(1.1xx10^(-7)-x)=10^(-14))

If we multiply this out we get a quadratic equation so the quadratic formula can be used to solve for sf(x). I won't go into that here but it gives, ignoring the -ve root:

sf(x=0.49xx10^(-8))

We can now get the equilibrium concentration of sf([H^+]):

sf([H^+]=10^(-7)-(0.49xx10^(-8))=9.51xx10^(-8)color(white)(x)"mol/l")

sf(pH=-log[H^+]=-log[9.51xx10^(-8)]=color(red)(7.02))

This is an example of "The Common Ion Effect", hydroxide being the common ion in question.

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