Question

How many gm of solid NaOH must be added to 100ml of a buffer solution which is 0.1M each w.r.t acid HA and salt NaA to make the pH of solution 5.5. Given `pK_a`(HA) = 5 (Use antilog (0.5) = 3.16?

Answer 1

`"0.2 g NaOH"`

Explanation:

You know that the hydroxide anions delivered to the solution by the sodium hydroxide will react with the weak acid to produce the conjugate base in `1:1` mole ratios.

`"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))`

The number of moles of weak acid and of conjugate base present in the buffer before the addition of the strong base is given by

`"moles HA" = "moles A"^(-) = "0.1 moles"/(10^3quad color(red)(cancel(color(black)("mL")))) * 100 color(red)(cancel(color(black)("mL")))`

`"moles HA = moles A"^(-) = "0.01 moles"`

Now, if you take `x` to be the number of moles of sodium hydroxide added to the buffer, you can say that after the reaction is complete, the resulting solution will contain

`(0.01 - x) quad "moles HA"`

The reaction will consume `x` moles of the weak acid.

`(0.01 + x) quad "moles A"^(-)`

The reaction will produce `x` moles of the conjugate base.

As you know, the `"pH"` of a weak acid-conjugate base buffer can be calculated using the Henderson - Hasselbalch equation.

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

Assuming that the volume of the buffer does not change upon the addition of the strong base, you can say that the volume of the solution after the strong base is added is equal to `"100 mL"`.

That said, the fact that the volume is the same for both the weak acid and the conjugate base allows you to treat the concentrations of the two chemical species and the number of moles interchangeably.

You can thus say that after the strong base is added, the `"pH"` of the solution will be equal to

`"pH" = "p"K_a + log [ ((0.01 + x) color(red)(cancel(color(black)("moles"))))/((0.01 - x) color(red)(cancel(color(black)("moles"))))]`

Plug in your values to find

`5.5 = 5 + log((0.01 +x)/(0.01 - x))`

This will be equivalent to

`log((0.01 + x)/(0.01 - x)) = 0.5`

To find the value of `x`, rewrite this as

`10^(log((0.01 + x)/(0.01-x))) = 10^(0.5)`

You will ned up with

`(0.01 + x)/(0.01 - x) = 3.16`

`0.01 + x = 0.0316 - 3.16x`

So

`x = (0.0316 - 0.01)/(1 + 3.16) = 0.00519`

Since `x` represents the number of moles of sodium hydroxide added to the buffer, you can use the molar mass of the compound to convert this to grams.

`0.0519 color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.2 g")))`

The answer is rounded to one significant figure, the number of sig figs you have for your values.