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#)?

Question

Serenity Allen · Accepted Answer

The pH of the buffer is 4.43.

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#

Cooper Adams · Answer

undefined To 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 (Ka = 1.74 × 10^-5 mol dm^-3), you need to use the Henderson-Hasselbalch equation:

pH = pKa + log([A^-]/[HA])

Where:
- pKa is the negative base-10 logarithm of the acid dissociation constant (Ka)
- [A^-] is the concentration of the conjugate base (acetate ion, CH3COO^-)
- [HA] is the concentration of the weak acid (ethanoic acid, CH3COOH)

First, calculate the concentrations of [A^-] and [HA]:
[A^-] = moles of ethanoic acid / total volume of solution
[HA] = moles of sodium hydroxide / total volume of solution

Next, calculate the pKa using the given Ka value.

Finally, substitute the calculated values into the Henderson-Hasselbalch equation and solve for pH.