Question

What is the pH of a 10 mL, 0.1 M NH3 solution when a 0.1 mL, 0.1 M NaOH solution was added into it?

I tried using an ICE table, but do not come up with a reasonable answer. Thank you in advance!

Answer 1

The `"pH = 11.3"`.

Explanation:

To calculate the new concentration of `"NaOH"` in the ammonia solution, you can use the dilution formula

`color(blue)(bar(ul(|color(white)(a/a) c_1V_1 = c_2V_2color(white)(a/a)|)))" "`

We can rearrange this formula to get

`c_2 = c_1 × V_1/V_2`

`c_1 = "0.1 mol/L"; V_1 = "0.1 mL"`
`c_2 = "?";color(white)(mmmml) V_2 = "(10 + 0.1) mL" = "10.1 mL"`

`c_2 = 0.1 "mol/L" × (0.10 color(red)(cancel(color(black)("mL"))))/(10.1 color(red)(cancel(color(black)("mL")))) = 9.9 ×10^"-4" "mol/L"`

Now, we can use an ICE table to solve the problem.

`color(white)(mmmmmm)"NH"_3 + "H"_2"O" → "NH"_4^"+" color(white)(m)+ color(white)(m)"OH"^"-"`
`"I/mol·L"^"-1":color(white)(m)0.1color(white)(mmmmmmm) 0color(white)(mmml)9.9×10^"-4"`
`"C/mol·L"^"-1":color(white)(ll) "-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmmmll) "+"x`
`"E/mol·L"^"-1":color(white)(l) "0.1 -" xcolor(white)(mmmmmm) x color(white)(mml)x + 9.9×10^"-4"`

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

`K_"b " = (x(x + 9.9×10^"-4"))/(0.1 - x) = 1.8 × 10^"-5"`

`0.1/(1.8 × 10^"-5") = 5600 > 400`

∴`x ≪ 0.1`

Then

`(x(x + 9.9×10^"-4"))/0.1 = 1.8 × 10^"-5"`

`x(x + 9.9×10^"-4") = 0.1 × 1.8 × 10^"-5"= 1.8 × 10^"-6"`

`x^2 + 9.9×10^"-4"x = 1.8 × 10^"-6"`

`x^2 + 9.9×10^"-4"x - 1.8 × 10^"-6" = 0`

`x = 9.9 × 10^"-4"`

∴ `["OH"^"-"] = (9.9 × 10^"-4" +x) color(white)(l)"mol/L" = (9.9 × 10^"-4" + 9.9 × 10^"-4") color(white)(l)"mol/L"`
`= 2.0 × 10^"-3"color(white)(l) "mol/L"`

`"pOH" = "-"log["OH"^"-"] = "-"log(2.0 ×10^"-3") = 2.7`

`"pH" = "14.00 - pOH" = "14.00 - 2.7" = 11.3`