Question

What is the pH of a `"10 mL"`, `"0.10 M NH"_3` solution after the a) addition of `"0.10 mL 1.00 M HCl"` b) of another `"0.10 mL 1.00 M HCl"` c) addition of `"0.10 mL 0.10 M NaOH"` and d) addition of another `"0.10 mL 0.10 M NaOH"`?

I calculated a pH of 10.22 for a) and a pH of 9.82 for b). They're so close it's not believable. Something might not be right with my ICE table. Help would be so very much appreciated. Thanks in advance!

Answer 1

WARNING! Very long answer!

Explanation:

a) After adding 0.10 mL `"HCl"`

`"Initial moles NH"_3 = 0.010 color(red)(cancel(color(black)("L NH"_3))) × "0.10 mol NH"_3/(1 color(red)(cancel(color(black)("L NH"_3)))) = "0.0010 mol NH"_3`

`"Moles HCl added" = "moles NH"_3color(white)(l) "reacted = moles NH"_4^+ "formed"`
`= "0.000 10" color(red)(cancel(color(black)("L"))) × "1.00 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.000 10 mol"`

`"Moles NH"_3color(white)(l) "remaining = (0.0010 - 0.000 10) mol NH"_3 = "0.0009 mol NH"_3`

`"NH"_3 + "H"_2"O" ⇌ "NH"_4^+ "+ OH"^"-"; "pK"_text(b)" = 4.75"`

`"B + H"_2"O" ⇌ "BH"^+ "+ OH"^"-"`

`"pOH" = "pK"_"b" + log((["BH"^+])/(["B"]))`

`"pOH" = 4.75 + log(("0.000 10")/0.0009) = 4.75 - 0.95 = 3.80`

`"pH = 14.00 - pOH = 14.00 - 3.80 = 10.20"`

b) After adding another 0.10 mol `"HCl"`

`"Total moles HCl added" = "moles NH"_3color(white)(l) "reacted = moles NH"_4^+ "formed" = "0.000 20 mol"`

`"Moles NH"_3 "remaining = (0.0010 - 0.000 20) mol NH"_3 = "0.0008 mol NH"_3`

`"pOH" = 4.75 + log((["BH"^+])/(["B"])) = 4.75 + log(0.0002/0.0008) = 4.75 - 0.60 = 4.15`

`"pH = 14.00 - pOH = 14.00 - 4.15 = 9.85"`

The `"pH"` of this solution should be close to that in Part a), because the solution in Part a) is a buffer.

c) After adding 0.10 mL NaOH

The strong base will dissociate completely.

`"Moles of NaOH" = "0.000 10" color(red)(cancel(color(black)("L NaOH"))) × "0.10 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = 1.0 × 10^"-5"color(white)(l) "mol"`

`V "= (10 + 0.10) mL = 10.1 mL = 0.0101 L"`

`["B"]_0 = "0.0010 mol"/"0.0101 L" = "0.099 mol/L"`

`["OH"^"-"]_0 =(1.0 × 10^"-5" "mol")/("0.0101 L") = 9.9 × 10^"-4"color(white)(l) "mol/L"`

`color(white)(mmmmmmmm)"B" + "H"_2"O" ⇌ "BH"^+ +color(white)(m) "OH"^"-"`
`"I/mol·L"^"-1":color(white)(mm) 0.099color(white)(mmmmml)0 color(white)(mmll)9.9 × 10^"-4"`
`"C/mol·L"^"-1":color(white)(mmll) "-"xcolor(white)(mmmmml)"+"xcolor(white)(mmmm)"+"x`
`"E/mol·L"^"-1":color(white)(ml)"0.099 -" x color(white)(mmmmll)xcolor(white)(ml)9.9 × 10^"-4" + x`

`K_"b" = (x(1.0 × 10^"-5" + x))/("0.099 -" x) = 10^"-4.75" = 1.78 × 10^"-5"`

`x` is negligible with respect to `0.099`, but not with respect to `9.9 × 10^"-4"`.

`(x(1.0 × 10^"-5" + x))/0.099 = 1.78 × 10^"-5"`

`1.0 × 10^"-5"x + x^2 = 1.78 × 10^"-5" × 0.099 = 1.76 × 10^"-8"`

`x^2 + 1.0 × 10^"-5"x - 1.76 × 10^"-8" = 0`

`x = 1.25 × 10^"-4"`

`["OH"^"-"] = 1.25 × 10^"-4" color(white)(l)"mol/L"`

`"pOH = -log"(1.25 × 10^"-4") = 3.90`

`"pH = 14.00 - 3.90 = 10.10"`

d) After adding another 0.10 mL `"NaOH"`

`"Moles of NaOH added" = 0.000 20 color(red)(cancel(color(black)("L NaOH"))) × "0.10 mol NaOH"/(1 color(red)(cancel(color(black)("L NaOH"))))`
`= 2.0 × 10^"-5"color(white)(l) "mol"`

`V "= (10 + 0.20) mL = 10.2 mL = 0.0102 L"`

`["B"]_0 = "0.0010 mol"/"0.0102 L" = "0.098 mol/L"`

`["OH"^"-"]_0 =(1.0 × 10^"-5" "mol")/("0.0102 L") = 9.8 × 10^"-4"color(white)(l) "mol/L"`

`color(white)(mmmmmmmm)"B" + "H"_2"O" ⇌ "BH"^+ +color(white)(m) "OH"^"-"`
`"I/mol·L"^"-1":color(white)(mm) 0.098color(white)(mmmmml)0 color(white)(mmll)9.8 × 10^"-4"`
`"C/mol·L"^"-1":color(white)(mmll) "-"xcolor(white)(mmmmml)"+"xcolor(white)(mmmm)"+"x`
`"E/mol·L"^"-1":color(white)(ml)"0.098 -" x color(white)(mmmmll)xcolor(white)(ml)9.8 × 10^"-4" + x`

`K_"b" = (x(1.0 × 10^"-5" + x))/0.098 = 10^"-4.75" = 1.78 × 10^"-5"`

`x` is negligible with respect to 0.098, but not with respect to `9.8 × 10^"-4"`.

`(x(1.0 × 10^"-5" + x))/0.098 = 1.78 × 10^"-5"`

`1.0 × 10^"-5"x + x^2 = 1.78 × 10^"-5" × 0.098 = 1.74 × 10^"-8"`

`x^2 + 1.0 × 10^"-5"x - 1.74 × 10^"-8" = 0`

`x = 1.25 × 10^"-4"`

`["OH"^"-"] = 1.25 × 10^"-4" color(white)(l)"mol/L"`

`"pOH = -log"(1.25 × 10^"-4") = 3.90`

`"pH = 14.00 - 3.90 = 10.10"`

In this case, adding the extra `"NaOH"` has no effect because its concentration is negligible compared with that provided by the `"NH"_3`.