# Calculate the change in pH for a buffer solution with an initial pH of #3.9# after addition of #"0.1 M HCl"#?

##### 1 Answer

I think you might be asking, "calculate the change in

Well, any properly made buffer would have the weak acid

It should go without saying that ** any concentration without the volume added is ambiguous**... Furthermore, you did not tell us what species are in solution, and in what concentrations...

So I will include three cases of an arbitrary acid and base at arbitrary concentrations... too bad for you I suppose?

- Some acid added, but not enough to get to half-equivalence point.
- Enough acid added to get to half equivalence point.
- Enough acid added to get to equivalence point.

As always, we use the Henderson-Hasselbalch equation when not yet at the equivalence point.

#"pH" = "pKa" + log\frac(["A"^(-)])(["HA"])#

Let's use formic acid for kicks...

#3.9 = 3.75 + log \frac(["A"^(-)])(["HA"])#

Currently we thus have

#\frac(["A"^(-)])(["HA"]) = 10^(3.9 - 3.75) = 1.413# , mostly weak base...

Let's suppose we therefore have a

**CASE I: NOT YET AT HALF EQUIV PT**

**We are still in the buffer region of the titration curve.**

Adding, arbitrarily,

#color(blue)("pH"') = 3.75 + log (("0.007063 mols" - "0.1 M" xx "0.005 L")/("0.005000 mol" + "0.1 M" xx "0.005 L"))#

#= color(blue)(ul(3.827))# ,as the mols of

#"HCl"# , denoted#n_(HCl)# , neutralize some mols of the conjugate base,#n_(A^(-))# , to generate equimolar quantities of the weak acid,#n_(HA)# .

**And we have not yet passed the half-equivalence point.**

**CASE II: AT HALF EQUIV PT**

This case is easy; at the **half-equivalence point**,

#color(blue)("pH"') = 3.75 + log((1)/(1)) = color(blue)(ul(3.75))#

So we might as well take it up a notch and calculate the volume of

#("0.007063 mols" - n_(HCl))/("0.005000 mols" + n_(HCl)) = 1#

#= 0.007063 - n_(HCl) = 0.005000 + n_(HCl)#

#=> n_(HCl) = 0.002063/2 = "0.001031 mols"#

And thus, the volume of

#V_(HCl) = "1000 mL"/(0.1 cancel"mols HCl") xx 0.001031 cancel"mols" = ul"10.314 mL"#

**CASE III: AT EQUIV PT**

**Here, we've completely neutralized all of the weak base and only have weak acid remaining.**

Thus, the Henderson-Hasselbalch equation *does NOT apply yet*, and we can only write the **dissociation reaction** for now:

#"HA"(aq) + "H"_2"O"(l) rightleftharpoons "A"^(-)(aq) + "H"_3"O"^(+)(aq)# with

#"pK"_a = 3.75# and#K_a = 1.778 xx 10^(-4)# .

At the start of this dissociation, we have a certain amount of

#["HA"]_i = (overbrace("0.005000 mols")^("mols HA") + overbrace("0.1 M" xx "0.07063 L")^"mols HCl added")/(underbrace("1 L")_"Soln volume" + underbrace("0.07063 L")_("HCl added")#

#=# #"0.01127 M"#

Then, the *mass action expression* is:

#K_a = 1.778 xx 10^(-4) = (["A"^(-)]["H"_3"O"^(+)])/(["HA"]) = x^2/(0.01127 - x)#

And from solving the quadratic equation for

#1.778 xx 10^(-4)(0.01127) - 1.778 xx 10^(-4)x - x^2 = 0# ,

we would get the physical answer,

This means the *current acid to base ratio* is:

#(["A"^(-)])/(["HA"]) = ("0.001329 M")/("0.01127 M" - "0.001329 M") = 0.1337#

And NOW, we can calculate the

**We are currently to the right of the equivalence point.**

#color(blue)("pH"') = 3.75 + log(0.1337)#

#= color(blue)(ul(2.876))#