Step 1. Calculate the moles of p-nitrophenol
#n =1.23color(red)(cancel(color(black)("g"))) × "1 mol"/(139.11 color(red)(cancel(color(black)("g")))) = "0.008 842 mol"#
Step 2. Calculate the molar concentration of p-nitrophenol
#c = ("0.008 842 mol")/"0.250 L" = "0.035 37 mol/L"#
Step 3. Solve the quadratic equation
We can use an ICE table here.
The chemical equation is
#"O"_2"NC"_6"H"_4"COOH" + "H"_2"O" ⇌ "H"_3"O"^"+" + "O"_2"NC"_6"H"_4"COO"^"-" ; "p"K_text(a) = 7.00"#
Let's rewrite this as
#color(white)(mmmmmmmmm)"HA" +color(white)(ll) "H"_2"O" ⇌ "H"_3"O"^"+" + "A"^"-"#
#"I/mol·L"^"-1":color(white)(mmll)"0.035 37"color(white)(mmmmmm)0color(white)(mmm)0#
#"C/mol·L"^"-1":color(white)(mmmll)"-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mm)"+"x#
#"E/mol·L"^"-1":color(white)(mm)"0.035 37 -"xcolor(white)(mmmmml)xcolor(white)(mmm)x#
#K_text(a) = 10^("-"7.00) = 1.00 × 10^"-7"#
#K_text(a) = (["H"_3"O"^"+"]["A"^"-"])/(["HA"]) = (x × x)/("0.035 37 -"color(white)(l)x) = x^2/("0.035 37 -"color(white)(l)x) = 1.00 × 10^"-7""#
Check for negligibility:
#"0.035 37"/(1.00 × 10^"-7") = 3.537 × 10^5 > 400#.
∴ #x ≪ "0.035 37"#
Then
#x^2/"0.035 37" = 1.00 × 10^"-7"#
#x^2= "0.035 37"× 1.00 × 10^"-7" = 3.537 ×10^"-9"#
#x = 5.95 × 10^"-5"#
Step 4. Calculate the pH
#["H"_3"O"^"+"] = x color(white)(l)"mol/L" = 5.95 × 10^"-5"color(white)(l)"mol/L"#
#"pH" = "-log"["H"_3"O"^"+"] = "-log"(5.95 × 10^"-5") = 4.22#
