Question #a4a33

The added water to reach #"100.00 mL"# doesn't change the mols of HCl present, but it does decrease the concentration by a factor of #100//40 = 2.5#. Regardless, what matters for neutralization is what amount of #"NaOH"# you add to what number of mols of #"HCl"#.

I got #"pH"#'s of #1.36, 1.51, 1.74, 2.54#.

You started with #"0.1100 M HCl"#, but it was diluted from #"40 mL"# to #"100 mL"#. That decreases its concentration by a factor of #2.5#.

#"0.1100 mol HCl"/"L" xx ("40 mL")/("100 mL")#

#= "0.04400 M HCl" = "0.04400 M H"^(+)#

since #"HCl"# is a strong acid.

For later, we will need that the starting mols of #"HCl"# is:

#"0.04400 mol/L HCl" xx 100.00 xx 10^(-3) "L"#

#= ul"0.004400 mols H"^(+)#

You're adding a strong base into it, so you just have to find how much #"H"^(+)# is leftover, or how much #"OH"^(-)# is leftover, after adding #"NaOH"# and neutralizing the #"H"^(+)# that was there from the #"HCl"# originally.

#a)#

Obviously, if you add no #"NaOH"#, you have not disturbed the #"HCl"#, so you can use the mols of #"HCl"# you have right now. You should then already know how much you have (this is your given).

#["H"^(+)]_1 = "0.04400 M"#

Thus,

#color(blue)("pH"_1) = -log["H"^(+)]_1#

#= -log(0.04400) = color(blue)(1.36)#

#b)#

Now you add #"10.00 mL"# of #"NaOH"#. This adds

#"0.1000 mol NaOH"/cancel"L" xx 10.00 xx 10^(-3) cancel"L"#

#= "0.001000 mols NaOH" = "0.001000 mols OH"^(-)#

This can then react as the limiting reactant with #"HCl"#. #"HCl"# has one proton, and #"NaOH"# has one #"OH"^(-)#, so they react #1:1#.

Thus, the mols of #"OH"^(-)# we just got are the mols of #"H"^(+)# lost.

#"mols H"^(+) = overbrace("0.004400 mols H"^(+))^("No NaOH") - overbrace("0.001000 mols OH"^(-))^("10.00 mL NaOH")#

#= "0.003400 mols H"^(+)#

and the new concentration of #"H"^(+)# is based on the new volume.

#["H"^(+)]_2 = ("0.003400 mols H"^(+))/((100.00 + 10.00) xx 10^(-3) "L")#

#= "0.03091 M H"^(+)#

and the #"pH"# is then

#color(blue)("pH"_2) = -log["H"^(+)]_2#

#= -log(0.03091) = color(blue)(1.51)#

#c)#

Hopefully you get the idea here. Now we add #"22.00 mL"# of #"NaOH"# INSTEAD of the #"10.00 mL"#. Our final volume then becomes #"122.00 mL"# instead of #"110.00 mL"#. The mols of #"OH"^(-)# added are:

#"0.1000 mol NaOH"/cancel"L" xx 22.00 xx 10^(-3) cancel"L"#

#= "0.002200 mols OH"^(-)#

And so, this neutralizes some #"H"^(+)# (consequently consuming all the #"OH"^(-)#) to give

#"mols H"^(+) = overbrace("0.004400 mols H"^(+))^("No NaOH") - overbrace("0.002200 mols OH"^(-))^("22.00 mL NaOH")#

#= "0.002200 mols H"^(+)#

and the new concentration of #"H"^(+)# is based on the new volume.

#["H"^(+)]_3 = ("0.002200 mols H"^(+))/((100.00 + 22.00) xx 10^(-3) "L")#

#= "0.01803 M H"^(+)#

and the #"pH"# is then

#color(blue)("pH"_3) = -log["H"^(+)]_3#

#= -log(0.01803) = color(blue)(1.74)#

At this point we have reached the "half-equivalence point", i.e. we have neutralized half of the #"HCl"#.

#d)#

Same thing here, but with #"40.00 mL"# added instead. The total volume is then #"140.00 mL"#, and the #"H"^(+)# is almost all gone.

#"0.1000 mol NaOH"/cancel"L" xx 40.00 xx 10^(-3) cancel"L"#

#= "0.004000 mols OH"^(-)#

And so, this neutralizes some #"H"^(+)# (consequently consuming all the #"OH"^(-)#) to give

#"mols H"^(+) = overbrace("0.004400 mols H"^(+))^("No NaOH") - overbrace("0.004000 mols OH"^(-))^("40.00 mL NaOH")#

#= "0.000400 mols H"^(+)#

and the new concentration of #"H"^(+)# is based on the new volume.

#["H"^(+)]_4 = ("0.000400 mols H"^(+))/((100.00 + 40.00) xx 10^(-3) "L")#

#= "0.00286 M H"^(+)#

and the #"pH"# is then

#color(blue)("pH"_4) = -log["H"^(+)]_4#

#= -log(0.00500) = color(blue)(2.54)#