The strategy to follow is
a) Write the chemical equation for the buffer.
b) Calculate the "pH" of the original buffer.
c) Calculate the moles of "HCl" added.
d) Calculate the new moles of "HA" and "A"^"-"
e) Calculate the "pH" of the new solution.
f) Calculate the change in "pH".
1) pH of Buffer
a). Chemical Equation
"HA" + "H"_2"O" ⇌ "H"_3"O"^(+) + "A"^"-"; K_text(a) = 6.3 × 10^"-4"
b) "pH" of original buffer
color(white)(XXXXm)"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^-; K_text(a) = 6.35 × 10^"-5"
"E/mol":color(white)(ll) 0.050 color(white)(XXXXXXXXXl)0.050
"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 "HA" and "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 "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
c) Moles of HCl added
Assume that you add 100 mL of 0.010 mol/L"HCl" to 100 mL of the buffer.
"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 "HA" and "A"^"-"
color(white)(XXXXXX)"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^-
"I/mol:"color(white)(XXl) 0.050color(white)(mmmmll)0.0010color(white)(Xl) 0.050
"C/mol:"color(white)(m)"+0.0010"color(white)(mmmll)"-0.0010"color(white)(m)"-0.0010"
"E/mol:"color(white)(XX)0.051color(white)(mmmmmm)0color(white)(XXll) 0.049
The added "H"_3"O"^"+" will increase the amount of "HA" and decrease the amount of "A"^"-" by 0.0010 mol.
e) "pH" of new solution
"pH" = "p"K_"a" + log(("[A"^(-)"]")/"[HA]") = 4.20 +log(0.049/0.051) = "4.20 - 0.017" = 4.183
f) Change in pH
"ΔpH = 4.20 – 4.183 = -0.017"
We usually express "pH" to only two decimal places, so
"ΔpH = -0.02"
