Calculate the pH of a #"0.2 M"# solution of #"NaC"_2"H"_3"O"_2# given that the hydrolysis constant is #K_h = 5.6xx10^(-10)#?

1 Answer
May 24, 2018

Answer:

pH = 9.02

Explanation:

We can use an ICE table to help with the calculation.

#color(white)(mmmmmmm)"A"^"-" + "H"_2"O" ⇌ "HA" + "OH"^"-"#
#"I/mol·L"^"-1":color(white)(mll)0.2color(white)(mmmmmm)0color(white)(mmll)0#
#"C/mol·L"^"-1":color(white)(mll)"-"xcolor(white)(mmmmml)"+"xcolor(white)(mm)"+"x#
#"E/mol·L"^"-1":color(white)(m)"0.2-"xcolor(white)(mmmmml)xcolor(white)(mmll)x#

#K_text(h) = (["HA"]["OH"^"-"])/(["A"^"-"]) = 5.6 × 10^"-10"#

#x^2/("0.2-"x) = 5.6 × 10^"-10"#

Apply the 5 % rule

#0.2/(5.6 × 10^"-10") = 3.6 × 10^8 > 400#.

#x ≪0.2#

Then

#x^2/0.2 = 5.6 × 10^"-10"#

#x^2 = 0.2(5.6 × 10^"-10") = 1.1 × 10^"-10"#

#x = 1.1 × 10^"-5"#

#["OH"^"-"] = x color(white)(l)"mol/L" = 1.1 × 10^"-5"color(white)(l) "mol/L"#

#"pOH" = "-log"["OH"^"-"] = "-log"(1.1 × 10^"-5") = 4.98#

#"pH = 14.00 - pOH = 14.00 - 4.98 = 9.02"#