Three beakers each contain #"100 mL"# of acidic solution with a #"pH"# of #3.00#. The acids in the beakers are #"HCl"#, #"HNO"_2#, and #"HI"#. If #"50 mL"# of #"0.1 M NaOH"# is added to each beaker, which resulting solution will have the lowest #"pH"#?

#"HCl"# and #"HI"#

The #"HCl"# and #"HI"# are both strong acids, so they will each give the same result.

Let's call them #"HX"#.

If

#"pH = 3.00"#

Then

#["H"_3"O"^"+"] = 10^"-3.00"color(white)(l)"mol/L" = 1.00 × 10^"-3"color(white)(l)"mol/L"#

The acid will react completely with the #"NaOH"#.

#"Moles of HX" = 100 color(red)(cancel(color(black)("mL HX"))) × (1.00 × 10^"-3" color(white)(l)"mmol HX")/(1 color(red)(cancel(color(black)("mL HX")))) = "0.100 mmol HX"#

#"Moles of NaOH" = 50 color(red)(cancel(color(black)("mL NaOH"))) × "0.1 mol NaOH"/(1 color(red)(cancel(color(black)("mL NaOH")))) = "5.0 mmol NaOH"#

#color(white)(mmmmmll)"HX + NaOH → NaX" + "H"_2"O"#
#"I/mol": color(white)(mll)0.100 color(white)(mm)5.0#
#"C/mol": color(white)(m)"-0.100"color(white)(ml)"-0.100"#
#"E/mol": color(white)(mll)0color(white)(mmmll)4.9#

So, we have 4.9 mmol of #"NaOH"# in 150 mL of solution.

#["OH"^"-"] = "4.9 mmol"/"150 mL" = "0.033 mol/L"#

#"pOH = -log"0.033 = 1.5#

#"pH = 14.00 - pOH = 14.00 - 1.5 = 12.5"#

#bb("HNO"_2)#

#"HNO"_2# is a weak acid with #K_text(a) = 4.0 × 10^"-4"#.

We must calculate the initial concentration of #"HNO"_2# that will give a final concentration of #1.00 × 10^"-3"color(white)(l)"mol/L H"_3"O"^"+"#.

#color(white)(mmmmmlmm)"HNO"_2 + "H"_2"O" ⇌ "H"_3"O"^"+"color(white)(m) +color(white)(mm) "NO"_2^"-"#
#"I/mol·l"^"-1": color(white)(mmmm)c color(white)(mmmmmmmll)0color(white)(mmmmmmll)0#
#"C/mol·l"^"-1": color(white)(m)"-1.00 × 10"^"-3"color(white)(mm)"+1.00 × 10"^"-3"color(white)(m)"+1.00 × 10"^"-3"#
#"E/mol·l"^"-1": color(white)(m)c"-1.00 × 10"^"-3"color(white)(mml)"1.00 × 10"^"-3"color(white)(mm)"1.00 × 10"^"-3"#

#K_text(a) = (["H"_3"O"^"+"]["NO"_2^"-"])/(["HNO"_2]) = (1.00 × 10^"-3")^2/(c - 1.00 ×10^"-3") = 4.0 × 10^"-4"#

#1.00 × 10^"-6" = 4.0 × 10^"-4"c - 4.00 × 10^"-7"#

#c = (1.00 × 10^"-6" + 4.00 × 10^"-7")/(4.0 × 10^"-4") = (1.40 × 10^"-6")/(4.0 × 10^"-4") = 3.5 × 10^"-3"#

#["HNO"_2] = 3.5 × 10^"-3"color(white)(l)"mol/L"#

The #"HNO"_2# will react completely with the #"NaOH"#.

#"Moles of HNO"_2 = 100 color(red)(cancel(color(black)("mL HNO"_2))) × (3.5 × 10^"-3" color(white)(l)"mmol HNO"_2)/(1 color(red)(cancel(color(black)("mL HNO"_2)))) = "0.35 mmol HX"#

#color(white)(mmmmll)"HNO"_2 + "NaOH → NaX" + "H"_2"O"#
#"I/mol": color(white)(mll)0.35 color(white)(mmm)5.0#
#"C/mol": color(white)(m)"-0.35"color(white)(mml)"-0.35"#
#"E/mol": color(white)(mll)0color(white)(mmmm)4.6#

So, we have 4.6 mmol of #"NaOH"# in 150 mL of solution.

#["OH"^"-"] = "4.6 mmol"/"150 mL" = "0.033 mol/L"#

#"pOH = -log"0.031 = 1.5#

#"pH = 14.00 - pOH = 14.00 - 1.5 = 12.5"#

All three solutions have the same pH.