Question

Consider a weak acid `"HA"`, where `"A"^(-)` is some arbitrary anion and `"HA"(aq) + "H"_2"O"(l) rightleftharpoons "H"_3"O"^(+)(aq) + "A"^(-)(aq)`. Suppose you add some `"NaA"(s)` to `"H"_2"O"(l)`. What is the pH of the solution?

Answer Choices
1. more than 7.00
2. less than 7.00
3. exactly 7.00

Answer 1

The overall pH would remain acidic but increase from pure water pH when in presence of common ion, `A^-`.

Explanation:

To understand this effect, one should calculate the %Ionization of a weak monoprotic acid in pure water and compare it to the %Ionization when the weak acid is in the presence of a quantity of `A^-.`

Assume the hypothetical problems...
(a) determine the %Ionization and pH of a 0.10M monoprotic acid (HA) in pure water. `K_a=1xx10^-5`, and
(b) determine the %Ionization and pH of a 0.10M monoprotic acid (HA) in a solution containing 0.01M NaA.

%Ionization of HA in pure water:
`color(white)(mmmmm)HA` `color(white)(m)rightleftharpoonscolor(white)(mm)H^+color(white)()+color(white)(m)A^-`
`C_i:color(white)(mm)0.10M` `color(white)(mmmm)0.00Mcolor(white)(m)0.00M`

`DeltaC:color(white)(m)-x` `color(white)(mmmmmm)+xcolor(white)(m)+x`

`C_(eq):color(white)(mm)(0.10-x)M` `color(white)(mm)xcolor(white)(mmm)x`
`color(white)(mmmm)~~0.10M`

`K_a=([H^+][A^-])/([HA])`=`(x^2)/(0.10)=1xx10^-5`

=> `x=[H^+]=sqrt((0.10)(1xx10^-5))=1xx10^-3M`
=> `%"Ionization"`=`([H^+])/([HA])xx100%=(10^-3)/(10^-1)xx100%=1.0%`
=> `pH=-log[H^+]=-log(10^-3)=3`

%Ionization of HA in 0.01M`A^-`:
`color(white)(mmmm)HA` `color(white)(mmm)rightleftharpoonscolor(white)(mmm)H^+color(white)()+color(white)(m)A^-`
`C_i:color(white)(m)0.10M` `color(white)(mmmmmm)0.00Mcolor(white)(mm)0.010M`

`DeltaC:color(white)(m)-x` `color(white)(mmmmmm)+xcolor(white)(mmmm)+x`

`C_(eq):color(white)(mm)(0.10-x)M` `color(white)(mm)xcolor(white)(mmm)(0.010+x)M`
`color(white)(mmmmm)~~0.10color(white)(mmmmmmmm)~~0.01M`

`K_a=([H^+][A^-])/([HA])=(x(0.01M))/(0.10M)=1xx10^-5`

=> `x=[H^+]=((0.1M)(1xx10^-5))/(0.01M)=1xx10^-4M`
=> `%"Ionization"`=`([H^+])/([HA])xx100%=(10^-4)/(10^-1)xx100%=0.10%`
=> `pH=-log[H^+]=-log(10^-4)=4`

So, in pure water pH HA => 3 but in presence of common ion (`A^-`) increases to 4 b/c of decreased ionization of HA. However, both systems remain pH < 7 (acidic).

Answer 2

The correct answer is 1. more than 7.00.

Explanation:

When you dissolve the weak acid `"HA"` in water, you get the equilibrium reaction.

`underbrace("HA(aq)")_color(red)("acid") + "H"_2"O(l)" ⇌ underbrace("A"^("-")"(aq)")_color(red)("base") + "H"_3"O"^("+")"(aq)"`

We could have written the equilibrium as

`underbrace("A"^("-")"(aq)")_color(red)("base") + "H"_3"O"^("+")"(aq)" ⇌ underbrace("HA(aq)")_color(red)("acid") + "H"_2"O(l)"`

The base removes a proton from the hydronium ion and forms water.

The base can remove a proton from water and form hydroxide ions.

`underbrace("A"^("-")"(aq)")_color(red)("base") + "H"_2"O(l)" ⇌ underbrace("HA(aq)")_color(red)("acid") + "OH"^"-""`

Thus, an aqueous solution of the salt of a weak acid is basic and the pH is greater than 7.

EXAMPLE

If the weak acid has `K_text(a) = 1.00 × 10^"-5"`, what is the pH of a 0.100 mol/L solution of `"NaA"`?

Solution

We can use an ICE table to help with the calculation.

`color(white)(mmmmmmmll)"A"^"-" + "H"_2"O" ⇌ "HA" + "OH"^"-"`
`"I/mol·L"^"-1":color(white)(mll)0.100color(white)(mmmmmll)0color(white)(mmll)0`
`"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mll)"+"x`
`"E/mol·L"^"-1":color(white)(m)"0.100-"xcolor(white)(mmmmm)xcolor(white)(mmm)x`

`K_text(b) = K_text(w)/K_text(a) = (1.00 × 10^"-14")/(1.00 × 10^"-5") = 1.00 × 10^"-9"`

`K_text(b) = (["HA"]["OH"^"-"])/(["A"^"-"]) = x^2/("0.100-"x) = 1.00 × 10^"-9"`

Apply the 5 % rule

`1.00/(1.00 × 10^"-9") = 1.00 × 10^9 > 400`.

∴ `x ≪1.00`

Then

`x^2/0.100 = 1.00 × 10^"-9"`

`x^2 = 0.100(1.00 × 10^"-9") = 1.00 × 10^"-10"`

`x = 1.00 × 10^"-5"`

`["OH"^"-"] = x color(white)(l)"mol/L" = 1.00 × 10^"-5"color(white)(l) "mol/L"`

`"pOH" = "-log"["OH"^"-"] = "-log"(1.00 × 10^"-5") = 5.00`

`"pH = 14.00 - pOH = 14.00 - 5.00 = 9.00"`