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"#
