How do I calculate the #"pH"# of the buffer solution formed when #10.0# #"cm"^3# of #0.80# #"mol"# #"dm"^-3# sodium hydroxide is mixed with #50.0# #"cm"^3# of #0.50# #"mol"# #"dm"^-3# ethanoic acid (#"K"_"a"=1.74xx10^-5# #"mol"# #"dm"^-3#)?

1 Answer
Apr 19, 2017

Answer:

The pH of the buffer is 4.43.

Explanation:

The first thing you must recognize is that there will be an acid-base neutralization reaction.

Your first task is to calculate the concentrations of the species present at the end of the reaction.

We can summarize the calculations in an ICE table.

#color(white)(mmmmml)"HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O"#
#"I/mol:"color(white)(mll)0.025color(white)(mm)0.0080color(white)(mm)0#
#"C/mol:"color(white)(m)"-0.0080"color(white)(m)"-0.0080"color(white)(m)"+0.0080"#
#"E/mol:"color(white)(ml)0.017color(white)(mm)0color(white)(mmmmll)0.0080#

#"Moles HA" = 0.0500 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA"))) × "0.50 mol HA"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"HA")))) = "0.025 mol HA"#

#"Moles OH"^"-" = 0.0100 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-"))) × "0.80 mol A"^"-"/(1 color(red)(cancel(color(black)("dm"^3 color(white)(l)"A"^"-")))) = "0.0080 mol A"^"-"#

So, at the end of the reaction, we have a solution containing 0.017 mol of #"HA"# and 0.0080 mol of #"A"^"-"#.

A mixture of a weak acid and its salt is a buffer.

We can use the Henderson-Hasselbalch equation to calculate the pH of the buffer:

#color(blue)(bar(ul(|color(white)(a/a)"pH" = "p"K_text(a) + log((["A"^"-"])/(["HA"]))color(white)(a/a)|)))" "#

Since both components are in the same solution, the ratio of their concentrations is the same as the ratio of their moles.

#"pH" = "-log"(1.74 × 10^"-5") + log((0.0080 color(red)(cancel(color(black)("mol"))))/(0.017 color(red)(cancel(color(black)("mol"))))) = "4.76 - 0.33" = 4.43#