How do you find the pH?

pKa = 10.20
Concentration of Acidic Solution = 0.8470

1 Answer
Apr 2, 2016

#"pH = 5.14"#

Explanation:

We know that #"p"K_"a" = 10.20#.

#K_"a" = 10^"-10.20" = 6.31 × 10^"-11"#

Let's set up an ICE table for the calculation of #["H"_3"O"^+]#.

#color(white)(mmmmmmm)"HA"color(white)(l) +color(white)(l) "H"_2"O"color(white)(l) ⇌ color(white)(l) "H"_3"O"^+color(white)(l) +color(white)(l) "A"^"-"#
#"I/mol·L"^"-1": color(white)(m)0.8470color(white)(mmmmmmm)0color(white)(mmmml)0#
#"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmmll)+xcolor(white)(mmll)+x#
#"E/mol·L"^"-1":color(white)(ll)"0.8470 -"xcolor(white)(mmmmmll)xcolor(white)(mmmml)x#

The #K_"a"# expression is

#K_"a" = (["H"_3"O"^+]["A"^"-"])/(["HA"]) = 6.31 × 10^"-11"#

#K_a = (x × x)/(0.8470-x) = x^2/(0.8470-x) = 6.31 × 10^"-11"#

Check if #x ≪ 0.8470#:

#0.8470/K_"a" = 0.8470/(6.31 × 10^"-11") = 1.34 × 10^10 ≫ 400#

#x ≪ 0.8470#, and the equation becomes

#x^2/0.8470 = 6.31 × 10^"-11"#

#x^2 = 0.8470 × 6.31 × 10^"-11" = 5.34 × 10^"-11"#

#x = 7.31 × 10^"-6"#

#["H"^+] = 7.31 × 10^-6 "mol/L"#

#"pH" = -log["H"^+] = -log(7.31 × 10^-6) = 5.14#