(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"H2PO-4+H2O⇌H3O++HPO2-4;Ka2=6.3×10-8
According to the Henderson-Hasselbalch Equation,
"pH" = "p"K_"a2" + log((["HPO"_4^"2-"])/(["H"_2"PO"_4^"-"]))pH=pKa2+log⎛⎜⎝[HPO2-4][H2PO-4]⎞⎟⎠
"p"K_"a2" = -log(6.3 × 10^-8) = 7.20pKa2=−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
