# Question 4c6d6

Jan 4, 2017

WARNING! Long answer! 1) $\text{pH = 4.20}$; 2) Δ"pH = -0.02"

#### Explanation:

a) Write the chemical equation for the buffer.
b) Calculate the $\text{pH}$ of the original 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) = 6.3 × 10^"-4"

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) = 6.35 × 10^"-5"
$\text{E/mol} : \textcolor{w h i t e}{l l} 0.050 \textcolor{w h i t e}{X X X X X X X X X l} 0.050$

$\text{Moles of HA" = "moles of A"^"-" = 0.050 color(red)(cancel(color(black)("L"))) × "1.0 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.050 mol}$

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(6.3 × 10^"-5") + log(0.050/0.050) = 4.20 + 0 = 4.20#

2) Change in pH

Assume that you add 100 mL of 0.010 mol/L$\text{HCl}$ to 100 mL of the buffer.

$\text{Moles of HCl" = "moles of H"_3"O"^"+" = 0.100 color(red)(cancel(color(black)("L"))) × "0.010 mol"/(1 color(red)(cancel(color(black)("L")))) = "0.0010 mol}$

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

$\textcolor{w h i t e}{X X X X X X} {\text{HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A}}^{-}$
$\text{I/mol:} \textcolor{w h i t e}{X X l} 0.050 \textcolor{w h i t e}{m m m m l l} 0.0010 \textcolor{w h i t e}{X l} 0.050$
$\text{C/mol:"color(white)(m)"+0.0010"color(white)(mmmll)"-0.0010"color(white)(m)"-0.0010}$
$\text{E/mol:} \textcolor{w h i t e}{X X} 0.051 \textcolor{w h i t e}{m m m m m m} 0 \textcolor{w h i t e}{X X l l} 0.049$

The added $\text{H"_3"O"^"+}$ will increase the amount of $\text{HA}$ and decrease the amount of $\text{A"^"-}$ by 0.0010 mol.

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

$\text{pH" = "p"K_"a" + log(("[A"^(-)"]")/"[HA]") = 4.20 +log(0.049/0.051) = "4.20 - 0.017} = 4.183$

f) Change in pH

$\text{ΔpH = 4.20 – 4.183 = -0.017}$

We usually express $\text{pH}$ to only two decimal places, so

$\text{ΔpH = -0.02}$