# Calculate the "pH" change in a buffer solution upon the addition of strong acids / base of known concentration?

## Calculate the $\text{pH}$ change that take place when a $100 \textcolor{w h i t e}{l} \text{ml}$ portion of A), B) is added to a $40 \textcolor{w h i t e}{l} \text{ml}$ of buffer solution made up of $0.2 \textcolor{w h i t e}{l} \text{M}$ ${\text{NH}}_{3}$ and $0.3 \textcolor{w h i t e}{l} \text{M}$ $\text{NH"_4"Cl}$ given that ${\text{Kb"=1.8 × 10^(−5) color(white)(l) "mol"\cdot "dm}}^{- 3}$ for ${\text{NH}}_{3}$. (Chemistry Libretext ) A) $0.5 \textcolor{w h i t e}{l} \text{M}$ $\text{NaOH}$ B) $0.05 \textcolor{w h i t e}{l} \text{M}$ $\text{HCl}$

Apr 7, 2018

A) $\Delta \text{pH} = + 3.50$
B) $\Delta \text{pH} = - 0.76$

#### Explanation:

Initial $\textsf{\text{pH}}$
Start by constructing a $\text{RICE}$ table (all figures are in ${\text{mol"\cdot "dm}}^{- 3}$)
for the dissociation of ${\text{NH}}_{3}$ (or $\text{NH"_3\cdot "H"_2"O} \left(a q\right)$ to be precise)
Let the decrease in the concentration of ammonia, $\text{NH"_3\cdot "H"_2"O} \left(a q\right)$, be $x \textcolor{w h i t e}{l} {\text{mol"\cdot "dm}}^{- 3}$.

Note that as a soluble ionic compound, $\text{NH"_4"Cl}$ ionizes completely to produce ${\text{NH}}_{4}^{+} \left(a q\right)$ and ${\text{Cl}}^{-} \left(a q\right)$ when dissolved in water.
Hence the concentration of ammonium ["NH"_4^(+)(aq)]=["NH"_4"Cl"]=0.3color(white)(l)"mol"\cdot"dm"^(-3) . The self-ionization of water is deemed neglectable.

$\textcolor{g r e y}{R} {\text{ " "NH"_3\cdot"H"_2"O" (aq)\rightleftharpoons "NH"_4^(+)(aq) + "OH}}^{-} \left(a q\right)$
$\textcolor{g r e y}{I} \textcolor{w h i t e}{- l l -} 0.008 \textcolor{w h i t e}{- - - - l l} 0.012$
$\textcolor{g r e y}{C} \textcolor{w h i t e}{- l l -} - x \textcolor{w h i t e}{- - l l - -} + x \textcolor{w h i t e}{- l -} + x$
$\textcolor{g r e y}{E} \textcolor{w h i t e}{- -} 0.008 - x \textcolor{w h i t e}{- l l -} 0.012 + x \textcolor{w h i t e}{- l l l} + x$

Here (in order to simplify the calculation) it is assumed that at equilibrium $0.2 - x \approx 0.2$ and $0.3 - x \approx 0.3$ given that the value of ${\text{K}}_{b}$ far smaller than that of either $\left[{\text{NH}}_{4}^{+}\right]$ or $\left[\text{NH"_3\cdot "H"_2"O}\right]$.

"K"_b=(["NH"_4^+]\cdot["OH"^-])/(["NH"_3\cdot "H"_2"O"])=(0.012\cdot x)/(0.008)

Solving for $x$ gives
$x = 1.2 \cdot {10}^{- 5} \textcolor{w h i t e}{l} {\text{mol"\cdot"dm}}^{- 3}$
Hence ["OH"^-]=1.2*10^(-5)color(white)(l)"mol"\cdot"dm"^(-3),

["H"^+]="K"_w/(["OH"^-])=8.3\cdot 10^(-10)color(white)(l)"mol"\cdot"dm"^(-3)
assuming that $\text{T} = 298 \textcolor{w h i t e}{l} K$ and "K"_w=1.0×10^(-14)

Therefore
ul("pH"_("initial")=-log["H"^+]=9.08)

$\text{pH}$ on the addition of $\textsf{100 \textcolor{w h i t e}{l} {\text{dm}}^{- 3}}$ of $\textsf{0.5 \textcolor{w h i t e}{l} {\text{mol"\cdot"dm}}^{- 3}}$ $\textsf{\text{NaOH}}$
Addition of sodium hydroxide changes the concentration of $\left[{\text{OH}}^{-}\right]$ in the initial solution; one might expect the solution to stay basic as it reaches equilibrium. All NH"_4^+ would have been converted to ${\text{NH}}_{3}$ with some ${\text{OH}}^{-}$ in excess. Thus it is possible to solve for the final $\left[{\text{OH}}^{-}\right]$ and hence $\text{pH}$ using the same $\text{RICE}$ table as in that the calculation of the initial $\text{pH}$.

["OH"^(-)(aq)]=["NaOH"]-["NH"_4^+(aq)]=0.038color(white)(l)"mol"

$\textcolor{g r e y}{R} {\text{ " "NH"_3\cdot"H"_2"O" (aq)\rightleftharpoons "NH"_4^(+)(aq) + "OH}}^{-} \left(a q\right)$
$\textcolor{g r e y}{I} \textcolor{w h i t e}{- l l -} 0.020 \textcolor{w h i t e}{- - - - - -} \textcolor{w h i t e}{- l l -} 0.038$
$\textcolor{g r e y}{C} \textcolor{w h i t e}{- l l -} - x \textcolor{w h i t e}{- - l l - -} + x \textcolor{w h i t e}{- l -} + x$
$\textcolor{g r e y}{E} \textcolor{w h i t e}{- l l} 0.020 - x \textcolor{w h i t e}{- - - -} x \textcolor{w h i t e}{- -} 0.038 + x$

Similarly,
$\frac{x \setminus \cdot 0.038}{0.020} = {\text{K}}_{b}$
$x = 6.32 \setminus \cdot {10}^{- 6}$

"pH"=14+log["OH"^-]=12.58

$\underline{\Delta \text{pH} = 3.50}$

$\text{pH}$ on the addition of $\textsf{100 \textcolor{w h i t e}{l} {\text{dm}}^{- 3}}$ of $\textsf{0.05 \textcolor{w h i t e}{l} {\text{mol"\cdot"dm}}^{- 3}}$ $\textsf{\text{HCl}}$
All $\left[{\text{H}}^{+}\right]$ were consumed in this case, converting part of ${\text{NH}}_{3}$ to ${\text{NH}}_{4}^{+}$. Similar to calculation of the initial $\text{pH}$,

$\textcolor{g r e y}{R} {\text{ " "NH"_3\cdot"H"_2"O" (aq)\rightleftharpoons "NH"_4^(+)(aq) + "OH}}^{-} \left(a q\right)$
$\textcolor{g r e y}{I} \textcolor{w h i t e}{- l l -} 0.003 \textcolor{w h i t e}{- - - - l l} 0.017$
$\textcolor{g r e y}{C} \textcolor{w h i t e}{- l l -} - x \textcolor{w h i t e}{- - l l - -} + x \textcolor{w h i t e}{- l -} + x$
$\textcolor{g r e y}{E} \textcolor{w h i t e}{- -} 0.003 - x \textcolor{w h i t e}{- l l -} 0.017 + x \textcolor{w h i t e}{- l l l} + x$

"K"_b=(["NH"_4^+]\cdot["OH"^-])/(["NH"_3\cdot "H"_2"O"])=(0.017\cdot x)/(0.003)

$x = 2.1 \cdot {10}^{- 6}$

"pH"=14+log["OH"^-]=8.32

$\underline{\Delta \text{pH} = - 0.76}$