Question #b7389

1 Answer
Apr 23, 2017

Answer:

Warning! Long Answer. pH = (a) 7.20; (b) 7.11; (c) 7.09

Explanation:

(a) pH of buffer

The equation for your buffer reaction is:

#"H"_2"PO"_4^"-" + "H"_2"O" ⇌ "H"_3"O"^"+" + "HPO"_4^"2-"; K_text(a2) = 6.3 × 10^"-8"#

According to the Henderson-Hasselbalch Equation,

#"pH" = "p"K_"a2" + log((["HPO"_4^"2-"])/(["H"_2"PO"_4^"-"]))#

#"p"K_"a2" = -log(6.3 × 10^-8) = 7.20#

#"pH" = 7.20 + log((0.50 cancel("mol/L"))/(0.50 cancel("mol/L"))) = 7.20 + 0 = 7.20#

The pH of the buffer is 7.20.

(b) Adding #"HCl"#

The buffer contains 0.50 mol #"HA"# and 0.50 mol #"A"^"-"#.

When we add 0.050 mol #"HCl"#, we remove 0.050 mol of #"A"^"-"# and form 0.050 mol of #"HA"#.

We can summarize the calculations in an ICE table.

#color(white)(mmmmml)"HA" color(white)(l)+ color(white)(m)"H"_3"O"^"+" → color(white)(ml)"A"^"-" + "H"_2"O"#
#"I/mol:"color(white)(mm)0.50color(white)(mml)0.050 color(white)(mmml)0.50#
#"C/mol:"color(white)(ll)"+0.050"color(white)(ml)"-0.050"color(white)(mmm)"-0.050"#
#"E/mol:"color(white)(mll)0.55color(white)(mmml)0color(white)(mmmml)0.45#

At the end of the reaction, we have a solution containing 0.55 mol of #"HA"# and 0.45 mol of #"A"^"-"#.

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" = 7.20 + log((0.45 color(red)(cancel(color(black)("mol"))))/(0.55 color(red)(cancel(color(black)("mol"))))) = "7.20 - 0.09" = 7.11#

(c) Adding#"NaOH"#

When we add 0.05 mol of #"NaOH"#, we remove 0.050 mol of #"HA"# and form 0.050 mol of #"A"^"-"#.

#color(white)(mmmmml)"HA" color(white)(l)+ color(white)(m)"OH"^"-" → color(white)(ml)"A"^"-" + "H"_2"O"#
#"I/mol:"color(white)(mm)0.50color(white)(mml)0.050 color(white)(mmll)0.50#
#"C/mol:"color(white)(m)"-0.050"color(white)(mll)"-0.050"color(white)(mll)"+0.050"#
#"E/mol:"color(white)(ml)0.45color(white)(mmmll)0color(white)(mmmll)0.55#

#"pH" = 7.20 + log((0.55 color(red)(cancel(color(black)("mol"))))/(0.45 color(red)(cancel(color(black)("mol"))))) = "7.20 + 0.09" = 7.29#