# 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!

Feb 6, 2017

The $\text{pH = 11.3}$.

#### Explanation:

To calculate the new concentration of $\text{NaOH}$ in the ammonia solution, you can use the dilution formula

$\textcolor{b l u e}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} {c}_{1} {V}_{1} = {c}_{2} {V}_{2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$

We can rearrange this formula to get

c_2 = c_1 × V_1/V_2

${c}_{1} = \text{0.1 mol/L"; V_1 = "0.1 mL}$
${c}_{2} = \text{?";color(white)(mmmml) V_2 = "(10 + 0.1) mL" = "10.1 mL}$

${c}_{2} = 0.1 \text{mol/L" × (0.10 color(red)(cancel(color(black)("mL"))))/(10.1 color(red)(cancel(color(black)("mL")))) = 9.9 ×10^"-4" "mol/L}$

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

$\textcolor{w h i t e}{m m m m m m} \text{NH"_3 + "H"_2"O" → "NH"_4^"+" color(white)(m)+ color(white)(m)"OH"^"-}$
$\text{I/mol·L"^"-1":color(white)(m)0.1color(white)(mmmmmmm) 0color(white)(mmml)9.9×10^"-4}$
$\text{C/mol·L"^"-1":color(white)(ll) "-"xcolor(white)(mmmmmmm)"+"xcolor(white)(mmmmll) "+} x$
$\text{E/mol·L"^"-1":color(white)(l) "0.1 -" xcolor(white)(mmmmmm) x color(white)(mml)x + 9.9×10^"-4}$

${K}_{\text{b" = (["NH"_4^"+"]["OH"^"-"])/(["NH"_3]) = 1.8 × 10^"-5}}$

${K}_{\text{b " = (x(x + 9.9×10^"-4"))/(0.1 - x) = 1.8 × 10^"-5}}$

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

$\text{pH" = "14.00 - pOH" = "14.00 - 2.7} = 11.3$