The strategy to follow is
a) Write the chemical equation for the buffer.
b) Calculate the "pH" of the 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) = 1.76 × 10^"-5"
b) "pH" of original buffer
color(white)(XXXXm)"HA" + "H"_2"O" ⇌ "H"_3"O"^+ + "A"^"-"; K_text(a) = 1.76 × 10^"-5"
"E/mol":color(white)(ll) 0.115 color(white)(XXXXXXXXXl)0.108
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(1.76 × 10^"-5") + log(0.108/0.115)
= 4.75 - 0.027 = 4.72
2) Change in pH on adding "HCl"
c) Moles of HCl added
You added 20.00 mL of 0.138 mol/L"HCl" to the buffer.
"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 "HA" and "A"^"-"
color(white)(XXXXXXll)"HA" + "H"_2"O" ⇌ "H"_3"O"^+color(white)(m) +color(white)(m) "A"^-
"I/mol:"color(white)(XXm) 0.115color(white)(mmmmll)"0.002 76"color(white)(Xm) 0.108
"C/mol:"color(white)(mll)"+0.00276"color(white)(mmml)"-0.002 76"color(white)(ml)"-0.00276"
"E/mol:"color(white)(XXll)0.1178color(white)(mmmmmm)0color(white)(XXmll) 0.1052
The added "H"_3"O"^"+" will increase the amount of "HA" and decrease the amount of "A"^"-" by 0.002 76 mol.
e) "pH" of new solution
"pH" = "p"K_"a" + log(("[A"^(-)"]")/"[HA]") = 4.75 + log(0.1052/0.1178) = "4.75 - 0.049" = 4.70
