Question

Find (a) the mols required to break the buffer (i.e. when `Delta"pH" = 1.00`), and (b) change in pH due to adding `"0.005 mols OH"^(-)` in a `"1-L"` acetic acid buffer containing `"0.010 M"` of each component? Then find the buffer capacity in (b).

Answer 1

(a) mols to break the buffer = `"0.008 mol"`; (b) `Delta"pH" = 0.35`
Buffer capacity = `"0.010 mol/L"cdot"pH"`

Explanation:

(a) Calculate the mols required

Buffer capacity is the amount of a monoprotic strong acid or base that must be added to 1 L of a buffer to change its pH by one unit.

Let's assume that we are adding `"NaOH"` to the buffer.

The base will decrease the amount of acetic acid and increase the amount of acetate.

Then we have

`color(white)(mmmmml)"HA + H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"`
`"I/mol":color(white)(mll)0.01color(white)(mmmmmmmmmll)0.01`
`"C/mol":color(white)(mll)"-"xcolor(white)(mmmmmmmmmmll)"+"x`
`"E/mol":color(white)(m)"0.01-"xcolor(white)(mmmmmmmml)"0.01+x"`

`"p"K_text(a) = 4.76`

The pH will increase to 5.76.

`"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))`

`5.76 = 4.76 + log((0.01+x)/(0.01-x))`

`1.00 = log((0.01+x)/(0.01-x))`

`(0.01+x)/(0.01-x) = 10^1.00 = 10.0`

`0.01+x =10.0(0.01-x) = 0.10-10.0x`

`11.0x = 0.09`

`x= 0.09/11.0 = 0.008`

The mols required is 0.008 mol.

(b) Calculate the change in pH on adding 0.005 mols of base

`color(white)(mmmmmmll)"HA + H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"`
`"I/mol·L"^"-1":color(white)(m)0.010color(white)(mmmmmmmmml)0.010`
`"C/mol·L"^"-1":color(white)(ll)"-0.005"color(white)(mmmmmmmll)"+0.005"`
`"E/mol·L"^"-1":color(white)(m)0.005color(white)(mmmmmmmmll)0.015`

For the original buffer,

`"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"])) = 4.76 + log(0.01/0.01) = 4.76 + log1 = 4.76 + 0 = 4.76`

After adding the base,

`"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))`

`"pH" = 4.76 + log(0.015/0.005) = 4.76 + log3 = 4.76 + 0.48 = 5.24`

`Δ"pH" = 5.24 - 4.76 = 0.48`

So, the buffer capacity at this point is:

`beta = (("0.005 mols OH"^(-))/("1 L buffer"))/("0.48 pH")`

`= "0.010 mol/L"cdot"pH"`