# Question b7389

Apr 23, 2017

Warning! Long Answer. pH = (a) 7.20; (b) 7.11; (c) 7.09

#### Explanation:

(a) pH of buffer

The equation for your buffer reaction is:

$\text{H"_2"PO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; K_text(a2) = 6.3 × 10^"-8}$

According to the Henderson-Hasselbalch Equation,

"pH" = "p"K_"a2" + log((["HPO"_4^"2-"])/(["H"_2"PO"_4^"-"]))

"p"K_"a2" = -log(6.3 × 10^-8) = 7.20

"pH" = 7.20 + log((0.50 cancel("mol/L"))/(0.50 cancel("mol/L"))) = 7.20 + 0 = 7.20

The pH of the buffer is 7.20.

(b) Adding $\text{HCl}$

The buffer contains 0.50 mol $\text{HA}$ and 0.50 mol $\text{A"^"-}$.

When we add 0.050 mol $\text{HCl}$, we remove 0.050 mol of $\text{A"^"-}$ and form 0.050 mol of $\text{HA}$.

We can summarize the calculations in an ICE table.

$\textcolor{w h i t e}{m m m m m l} \text{HA" color(white)(l)+ color(white)(m)"H"_3"O"^"+" → color(white)(ml)"A"^"-" + "H"_2"O}$
$\text{I/mol:} \textcolor{w h i t e}{m m} 0.50 \textcolor{w h i t e}{m m l} 0.050 \textcolor{w h i t e}{m m m l} 0.50$
$\text{C/mol:"color(white)(ll)"+0.050"color(white)(ml)"-0.050"color(white)(mmm)"-0.050}$
$\text{E/mol:} \textcolor{w h i t e}{m l l} 0.55 \textcolor{w h i t e}{m m m l} 0 \textcolor{w h i t e}{m m m m l} 0.45$

At the end of the reaction, we have a solution containing 0.55 mol of $\text{HA}$ and 0.45 mol of $\text{A"^"-}$.

We can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:

color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))color(white)(a/a)|)))" "#

Since both components are in the same solution, the ratio of their concentrations is the same as the ratio of their moles.

$\text{pH" = 7.20 + log((0.45 color(red)(cancel(color(black)("mol"))))/(0.55 color(red)(cancel(color(black)("mol"))))) = "7.20 - 0.09} = 7.11$

(c) Adding$\text{NaOH}$

When we add 0.05 mol of $\text{NaOH}$, we remove 0.050 mol of $\text{HA}$ and form 0.050 mol of $\text{A"^"-}$.

$\textcolor{w h i t e}{m m m m m l} \text{HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O}$
$\text{I/mol:} \textcolor{w h i t e}{m m} 0.50 \textcolor{w h i t e}{m m l} 0.050 \textcolor{w h i t e}{m m l l} 0.50$
$\text{C/mol:"color(white)(m)"-0.050"color(white)(mll)"-0.050"color(white)(mll)"+0.050}$
$\text{E/mol:} \textcolor{w h i t e}{m l} 0.45 \textcolor{w h i t e}{m m m l l} 0 \textcolor{w h i t e}{m m m l l} 0.55$

$\text{pH" = 7.20 + log((0.55 color(red)(cancel(color(black)("mol"))))/(0.45 color(red)(cancel(color(black)("mol"))))) = "7.20 + 0.09} = 7.29$