Question

If 1000 molecules of HC2H3O2 are placed into solution. How many hydronium ions will form? 1 5 10 100

Answer 1

Should you not quote the acid dissociation constant....?

Explanation:

I would suspect ONE hydronium ion to result if that....in the given scenario....

Answer 2

There will be 10 hydronium ions.

Explanation:

The degree of ionization depends on the concentration.

You don't state the concentration of the acetic acid, so I will assume that it is 0.200 mol/L.

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)0.200color(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)"0.200-"xcolor(white)(mmmmll)xcolor(white)(mmll)x`

`K_text(a) = (["A"^"-"]["H"_3"O"^"+"])/(["HA"]) = x^2/(00-x) = 1.76 × 10^"-5"`

Check for negligibility:

`0.200/(1.76 ×10^"-5") = 11 000 > 400`. ∴ x ≪ 0.200.

`x^2/0.200 = 1.76 × 10^"-5"`

`x^2 = 0.200 × 1.76 × 10^"-5" = 3.52 × 10^"-6"`

`x = 1.88 × 10^"-3"`

`["H"_3"O"^"+"] = xcolor(white)(l)"mol/L" = 1.88 × 10^"-3"color(white)(l)"mol/L"`

`"Fraction ionized" = (["H"_3"O"^"+"])/(["HA"]_0) = (1.88 × 10^"-3")/0.200 = 0.009`

This result means that 9 molecules out of every thousand are in the ionized form.