# Question 469ae

Jan 23, 2017

$\text{Old pH = 4.72}$; $\text{New pH = 4.70}$

#### Explanation:

a) Write the chemical equation for the buffer.
b) Calculate the $\text{pH}$ of the buffer.

c) Calculate the moles of $\text{HCl}$ added.
d) Calculate the new moles of $\text{HA}$ and $\text{A"^"-}$
e) Calculate the $\text{pH}$ of the new solution.
f) Calculate the change in $\text{pH}$.

1) pH of Buffer

a). Chemical Equation

$\text{HA" + "H"_2"O" ⇌ "H"_3"O"^(+) + "A"^"-}$; K_text(a) = 1.76 × 10^"-5"

b) $\text{pH}$ of original buffer

$\textcolor{w h i t e}{X X X X m} \text{HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-}$; K_text(a) = 1.76 × 10^"-5"
$\text{E/mol} : \textcolor{w h i t e}{l l} 0.115 \textcolor{w h i t e}{X X X X X X X X X l} 0.108$

Since both $\text{HA}$ and $\text{A"^"-}$ are in the same solution, the ratio of their moles is the same as the ratio of their molarities.

We can use the Henderson-Hasselbalch equation to calculate the $\text{pH}$.

"pH" = "p"K_"a" + log(("[A"^"-""]")/"[HA]") = -log(1.76 × 10^"-5") + log(0.108/0.115)#
$= 4.75 - 0.027 = 4.72$

2) Change in pH on adding $\text{HCl}$

You added 20.00 mL of 0.138 mol/L$\text{HCl}$ to the buffer.

$\text{Moles of HCl added " = "moles of H"_3"O"^"+" = "0.020 00" color(red)(cancel(color(black)("L"))) × "0.138 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.002 76 mol}$

d) New moles of $\text{HA}$ and $\text{A"^"-}$

$\textcolor{w h i t e}{X X X X X X l l} {\text{HA" + "H"_2"O" ⇌ "H"_3"O"^+color(white)(m) +color(white)(m) "A}}^{-}$
$\text{I/mol:"color(white)(XXm) 0.115color(white)(mmmmll)"0.002 76} \textcolor{w h i t e}{X m} 0.108$
$\text{C/mol:"color(white)(mll)"+0.00276"color(white)(mmml)"-0.002 76"color(white)(ml)"-0.00276}$
$\text{E/mol:} \textcolor{w h i t e}{X X l l} 0.1178 \textcolor{w h i t e}{m m m m m m} 0 \textcolor{w h i t e}{X X m l l} 0.1052$

The added $\text{H"_3"O"^"+}$ will increase the amount of $\text{HA}$ and decrease the amount of $\text{A"^"-}$ by 0.002 76 mol.

e) $\text{pH}$ of new solution

$\text{pH" = "p"K_"a" + log(("[A"^(-)"]")/"[HA]") = 4.75 + log(0.1052/0.1178) = "4.75 - 0.049} = 4.70$