# 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!

Sep 30, 2016

#### Explanation:

a) After adding 0.10 mL $\text{HCl}$

${\text{Initial moles NH"_3 = 0.010 color(red)(cancel(color(black)("L NH"_3))) × "0.10 mol NH"_3/(1 color(red)(cancel(color(black)("L NH"_3)))) = "0.0010 mol NH}}_{3}$

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

${\text{Moles NH"_3color(white)(l) "remaining = (0.0010 - 0.000 10) mol NH"_3 = "0.0009 mol NH}}_{3}$

$\text{NH"_3 + "H"_2"O" ⇌ "NH"_4^+ "+ OH"^"-"; "pK"_text(b)" = 4.75}$

$\text{B + H"_2"O" ⇌ "BH"^+ "+ OH"^"-}$

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

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

$\text{pH = 14.00 - pOH = 14.00 - 3.80 = 10.20}$

b) After adding another 0.10 mol $\text{HCl}$

$\text{Total moles HCl added" = "moles NH"_3color(white)(l) "reacted = moles NH"_4^+ "formed" = "0.000 20 mol}$

${\text{Moles NH"_3 "remaining = (0.0010 - 0.000 20) mol NH"_3 = "0.0008 mol NH}}_{3}$

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

$\text{pH = 14.00 - pOH = 14.00 - 4.15 = 9.85}$

The $\text{pH}$ 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.

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

$V \text{= (10 + 0.10) mL = 10.1 mL = 0.0101 L}$

["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"

$\textcolor{w h i t e}{m m m m m m m m} \text{B" + "H"_2"O" ⇌ "BH"^+ +color(white)(m) "OH"^"-}$
$\text{I/mol·L"^"-1":color(white)(mm) 0.099color(white)(mmmmml)0 color(white)(mmll)9.9 × 10^"-4}$
$\text{C/mol·L"^"-1":color(white)(mmll) "-"xcolor(white)(mmmmml)"+"xcolor(white)(mmmm)"+} x$
$\text{E/mol·L"^"-1":color(white)(ml)"0.099 -" x color(white)(mmmmll)xcolor(white)(ml)9.9 × 10^"-4} + x$

${K}_{\text{b" = (x(1.0 × 10^"-5" + x))/("0.099 -" x) = 10^"-4.75" = 1.78 × 10^"-5}}$

$x$ 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

$\text{pH = 14.00 - 3.90 = 10.10}$

d) After adding another 0.10 mL $\text{NaOH}$

"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"

$V \text{= (10 + 0.20) mL = 10.2 mL = 0.0102 L}$

["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"

$\textcolor{w h i t e}{m m m m m m m m} \text{B" + "H"_2"O" ⇌ "BH"^+ +color(white)(m) "OH"^"-}$
$\text{I/mol·L"^"-1":color(white)(mm) 0.098color(white)(mmmmml)0 color(white)(mmll)9.8 × 10^"-4}$
$\text{C/mol·L"^"-1":color(white)(mmll) "-"xcolor(white)(mmmmml)"+"xcolor(white)(mmmm)"+} x$
$\text{E/mol·L"^"-1":color(white)(ml)"0.098 -" x color(white)(mmmmll)xcolor(white)(ml)9.8 × 10^"-4} + x$

${K}_{\text{b" = (x(1.0 × 10^"-5" + x))/0.098 = 10^"-4.75" = 1.78 × 10^"-5}}$

$x$ 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

$\text{pH = 14.00 - 3.90 = 10.10}$

In this case, adding the extra $\text{NaOH}$ has no effect because its concentration is negligible compared with that provided by the ${\text{NH}}_{3}$.