Question

45 ml of 1.89E-2 M HC2H3O2 is prepared at 25 C ka is 1.75E-5 1. What is the percent ionization of the acid? pH? 2. What is the percent ionization of the acid and pH if 1.8 grams of NaC2H3O2 is added to the solution?

I am not sure how to even properly start this question.

I initially started by multiplying the molarity of HC2H3O2 by .045 Lto get the answer in moles and created an ice table based on this equation

C2H3O2 (s) + H2O (l)--> HC2H3O2(aq)+ +OH-(aq)

but I can't end up with an answer with moles. Idk how to do this

Answer 1

Warning! Long Answer.
1. `"% Ionization = 2.0 %";color(white)(mm)"pH = 3.06"`.
2. `"% Ionization = 0. 0036 %; pH = 5.79"`.

Explanation:

1. Percent ionization

We can use an ICE table to solve the problem.

`color(white)(mmmmmmml)"HA + H"_2"O" ⇌ "A"^"-" + "H"_3"O"^"+"`
`"I/mol·L"^"-1": color(white)(mll)0.045 color(white)(mmmmml)0color(white)(mmll)0`
`"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmll)"+"xcolor(white)(mll)"+"x`
`"E/mol·L"^"-1": color(white)(m)"0.045-"xcolor(white)(mmmmll)xcolor(white)(mmll)x`

`K_text(a) = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = x^2/(0.045-x) = 1.75 × 10^"-5"`

Check for negligibility:

`0.045/(1.75 ×10^"-5") = 2600 > 400`. ∴ x ≪ 0.045.

`x^2/0.045 = 1.75 × 10^"-5"`

`x^2 = 0.045 × 1.75 × 10^"-5" = 7.88 × 10^"-7"`

`x = 8.78× 10^"-4"`

`"% Ionization" = x/(["HA"]_0) × 100 % = (8.78× 10^"-4")/0.045 × 100 % = 2.0 %`

`["H"_3"O"^"+"] = xcolor(white)(l)"mol/L" = 8.78 × 10^"-4"color(white)(l)"mol/L"`

`"pH = -log"["H"_3"O"^"+"] = "-log"(8.78 × 10^"-4") = 3.06`

2. Percent ionization of new solution

(a) Calculate the moles of sodium acetate

`n = 1.8 color(red)(cancel(color(black)("g NaC"_2"H"_3"O"_2))) × ("1 mol NaC"_2"H"_3"O"_2)/(82.03 color(red)(cancel(color(black)("g NaC"_2"H"_3"O"_2)))) = "0.0219 mol NaC"_2"H"_3"O"_2`

(b) Calculate the concentration of sodium acetate

`["A"^"-"] = "21.9 mmol"/"45 mL" = "0.488 mol/L"`

(c) Calculate the new percent ionization

Per Le Châtelier's Principle, the addition of a common ion (acetate ion) will decrease the percent ionization.

`color(white)(mmmmmmml)"HA + H"_2"O" ⇌ "H"_3"O"^"+" +color(white)(ll)"A"^"-"`
`"I/mol·L"^"-1": color(white)(mll)0.045 color(white)(mmmmml)0color(white)(mmml)0.488`
`"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmmmll)"+"xcolor(white)(mmml)"+"x`
`"E/mol·L"^"-1": color(white)(m)"0.045-"xcolor(white)(mmmmll)xcolor(white)(mmll)"0.488+"x`

`K_text(a) = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = (x(0.488+x))/(0.045-x) = 1.75 × 10^"-5"`

We have already determined that `x ≪ 0.045`.

`(x× 0.48)/0.045 = 1.75 × 10^"-5"`

`x = (0.045 × 1.75 × 10^"-5")/0.48 = 7.88 × 10^"-7"`

`x = 1.61 × 10^"-6"`

`"% Ionization" = x/(["HA"]_0) × 100 % = (1.61 × 10^"-6")/0.045 × 100 % = 0.0036 %`

(d) Calculate the pH

`["H"_3"O"^"+"] = xcolor(white)(l)"mol/L" = 1.61 × 10^"-6"color(white)(l)"mol/L"`

`"pH = -log"["H"_3"O"^"+"] = "-log"(1.61 × 10^"-6") = 5.79`