Question

Calculate the pH of a solution labelled 0.62 M ammonia, NH3(aq), which is a weak monoprotic base. Kb(NH3) = 1.8 × 10−5. Use the 5% approximation rule. Provide your answer to two places after the decimal.?

Answer 1

pH =11.52

Explanation:

We can use an ICE table to help with the calculation.

`color(white)(mmmmmmm)"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"`
`"I/mol·L"^"-1":color(white)(mll)0.62color(white)(mmmmmmll)0color(white)(mmm)0`
`"C/mol·L"^"-1":color(white)(mm)"-"xcolor(white)(mmmmmml)"+"xcolor(white)(mm)"+"x`
`"E/mol·L"^"-1":color(white)(m)"0.62-"xcolor(white)(mmmmmm)xcolor(white)(mmm)x`

`K_text(b) = (["NH"_4^"+"]["OH"^"-"])/(["NH"_3]) = 1.8 × 10^"-5"`

`x^2/("0.62-"x) = 1.8 × 10^"-5"`

Appy 5 % rule

`0.62/(1.8 × 10^"-5") = "34 000" > 400`.

∴ `x ≪0.62`

Then

`x^2/0.62 = 1.8 × 10^"-5"`

`x^2 = 0.62(1.8 × 10^"-5") = 1.12 × 10^"-5"`

`x = 3.34 × 10^"-3"`

`["OH"^"-"] = x color(white)(l)"mol/L" = 3.34 × 10^"-3"color(white)(l) "mol/L"`

`"pOH" = "-log"["OH"^"-"] = "-log"(3.34 × 10^"-3") = 2.48`

`"pH = 14.00 - pOH = 14.00 - 2.48 = 11.52"`