A 1.0 M solution of a weak acid, HB, has an [H3O+] equal to 0.025 M. Find the value of Ka. (HB + H2O --> B- + H3O+) Can anyone help?

1 Answer
May 2, 2018

# K_text(a) = 6.4 × 10^"-4"#

Explanation:

We can use an ICE table to help us solve this problem.

#color(white)(mmmmmmll)"HB" + "H"_2"O" ⇌ "H"_3"O"^"+" + color(white)(l)"B"^"-"#
#"I/mol·L"^"-1": color(white)(mll)1.0 color(white)(mmmmmmll)0color(white)(mmml)0#
#"C/mol·L"^"-1": color(white)(ml)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mml)"+"x#
#"E/mol·L"^"-1": color(white)(m)"1.0-"xcolor(white)(mmmmmm)xcolor(white)(mmml)x#

At equilibrium,

#["H"_3"O"^"+"] = "0.025 mol/L"#

#x = 0.025#

#K_text(a) = (["H"_3"O"^"+"]["B"^"-"])/(["HB"]) = x^2/(1.0-x) = 0.025^2/(1.0-0.025) = (6.25 × 10^"-4")/0.975 = 6.4 × 10^"-4"#