Question

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)`?

Answer 1

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"`