Question #7bd01

1 Answer
Oct 9, 2017

Answer:

WARNING! Long answer! (a) pH = 3; (b) The buffer must be diluted 820-fold.

Explanation:

(a) pH of the buffer

We can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:

#color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))color(white)(a/a)|)))" "#

#"pH" = 3 + log((0.10 color(red)(cancel(color(black)("mol/L"))))/(0.10 color(red)(cancel(color(black)("mol/L"))))) = 3 + log(1.0) =3 + 0.00 = 3#

(b) Dilution to pH 4

Dilution by, say, a factor of 10 has no effect on the pH of the buffer.

However, if you make the buffer extremely dilute, the pH will approach that of pure water (pH 7.00).

One way to approach the problem is to assume that we start with pure water and add equal concentrations of the acid and its salt.

Then, we calculate the what those concentrations must be to get pH = 4.

They will be the concentrations to which we must dilute the original buffer.

So, we start with pure water containing #xcolor(white)(l)"mol/L"# of #"HA"# and #"A"^"-""#.

#color(white)(mmmmmmmmll)"HA" + "H"_2"O" ⇌ color(white)(ml)"H"_3"O"^"+"color(white)(m) +color(white)(mmll) "A"^"-"#
#"I/mol·L"^"-1":color(white)(mmmm)x color(white)(mmmmmml)1.00 × 10^"-7"color(white)(mmmm)x#
#"C/mol·L"^"-1": color(white)(ml)"-9.99" × 10^"-5"color(white)(mmll) "+9.99" × 10^"-5"color(white)(mll) "+9.99" × 10^"-5"#
#"E/mol·L"^"-1": color(white)(m)x - 9.99 × 10^"-5"color(white)(mlm) 1.00 × 10^"-4"color(white)(m)x + 9.99 × 10^"-5"#

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

#4 = 3 + log((x + 9.99 × 10^"-5")/(x - 9.99 × 10^"-5"))#

#1 = log((x + 9.99 × 10^"-5")/(x - 1.99 × 10^"-5"))#

#10 = (x+ 9.99 × 10^"-5")/(x -9.99 × 10^"-5")#

#10x - 9.99 × 10^"-4" = x + 9.99 × 10^"-5"#

#9x = "9.99" × 10^"-5" + 9.99 × 10^"-4" = 1.10 × 10^"-3"#

#x = 1.2 × 10^"-4"#

#"Dilution factor" = (0.10 color(red)(cancel(color(black)("mol/L"))))/(1.2 × 10^"-4" color(red)(cancel(color(black)("mol/L")))) = 820#

The buffer must be diluted 180-fold to make the formal concentrations of the acid and its salt equal #1.2 × 10^"-4"color(white)(l)"mol/L"#

Check

#"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"])) = 3 + log((1.2 × 10^"-4" + 9.99 × 10^"-5")/(1.2 × 10^"-4" - 9.99 × 10^"-5"))#

#= 3 + log((2.2 × 10^"-4")/(2.2 × 10^"-5")) = 3 + log10 = 3 + 1 = 4#

It checks!