Question #469ae

1 Answer
Jan 23, 2017

Answer:

#"Old pH = 4.72"#; #"New pH = 4.70"#

Explanation:

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#