What is the Ka of an acid of 6M which has a pH of 2.5 at 277.5K ?

1 Answer
Mar 17, 2017

Answer:

#K_text(a) = 2 × 10^"-6"#

Explanation:

I would argue that the temperature has nothing to do with the value of #K_text(a)#.

In this problem, it is the #"pH"# and the initial concentration of the acid that determine the value of #K_text(a)#.

Let's set up an ICE table to solve this problem.

#color(white)(mmmmmmm)"HA" + "H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"#
#"I/mol·L"^"-1":color(white)(mml)6color(white)(mmmmmml)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmm)"+"xcolor(white)(mml)"+"x#
#"E/mol·L"^"-1":color(white)(ml)"6 -"color(white)(l)xcolor(white)(mmmmml)xcolor(white)(mmll)x#

We must use the #"pH"# to calculate the value of #x#.

#"pH = 2.5"#

#["H"_3"O"^"+"] = 10^"-2.5"color(white)(l) "mol/L" = "0.0031 mol/L" = x#

The #K_"a"# expression is:

#K_"a" = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x × x)/(0.100-x) = x^2/(0.100-x) = 0.0031^2/(6 - 0.0031) = (1.00 × 10^"-5")/6 = 2 × 10^"-6"#