What is the pH of a solution prepared by dissolving 1.23 g of 2-nitrophenol in 0.250L? pKa=7.00

1 Answer
Jan 12, 2018

pH = 4.22

Explanation:

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