This question has so many parts that I am not giving detailed explanations.
a- The chemical equation
#"NH"_3 + "HCl" → "NH"_4"Cl"#
b- Volume of #"HCl"# at the equivalence point
#"Moles of NH"_3 = 0.050 color(red)(cancel(color(black)("L NH"_3))) × "0.1 mol NH"_3/(1 color(red)(cancel(color(black)("L NH"_3)))) = "0.005 mol NH"_3#
#"Moles of HCl" = 0.005 color(red)(cancel(color(black)("mol NH"_3))) × "1 mol HCl"/(1 color(red)(cancel(color(black)("mol NH"_3)))) = "0.005 mol HCl"#
#"Volume of HCl" = 0.005 color(red)(cancel(color(black)("mol HCl"))) × "1 L HCl"/(0.1 color(red)(cancel(color(black)("mol HCl")))) = "0.05 L HCl" = "50 mL HCl"#
c- pH at different volumes
(i) pH at 0 mL
#color(white)(mmmmmml)"NH"_3 + "H"_2"O" ⇌ "NH"_4^"+" + "OH"^"-"#
#"I/mol·L"^"-1": color(white)(mll)0.1color(white)(mmmmmmll)0color(white)(mmm)0#
#"C/mol·L"^"-1": color(white)(mll)"-"xcolor(white)(mmmmmm)"+"xcolor(white)(mml)"+"x#
#"E/mol·L"^"-1": color(white)(m)"0.1-"xcolor(white)(mmmmmm)xcolor(white)(mmm)x#
#K_text(b) = (["NH"_4^"+"]["OH"^"-"])/(["NH"_3]) = x^2/(0.1-x) = 1.76 × 10^"-5"#
#x^2 = 0.1 × 1.76 × 10^"-5" = 1.8 × 10^"-6"#
#x = 1.3 × 10^"-3"#
#["OH"^"-"] = xcolor(white)(l) "mol/L" = 1.3 × 10^"-3" color(white)(l)"mol/L"#
#"pOH" = "-log"(1.3 × 10^"-3") = 2.88#
#"pH = 14.00 - 2.88 = 11.12"#
(ii) pH at 10 mL #"HCl"#
#color(white)(mmmmml)"NH"_3 + "H"_3"O"^"+" → "NH"_4^"+"#
#"I/mol": color(white)(mll)0.005color(white)(ml)0.001color(white)(mmm)0#
#"C/mol": color(white)(m)"-0.001"color(white)(m)"-0.001"color(white)(m)"+0.001"#
#"E/mol": color(white)(ml)0.004color(white)(mml)0color(white)(mmm)0.001#
This is a buffer!
#"p"K_text(b) = "-log"(1.76 × 10^"-5") = 4.75#
#"pOH" = "p"K_text(b) + log((["NH"_4^"+"])/(["NH"_3])) = 4.75 + log(0.001/0.004) = 4.75 - 0.60 = 4.15#
#"pH" = 14.00 - 4.15 = 9.85"#
(iii) pH at the equivalence point.
We have neutralized all the #"NH"_3#, so we have 0.005 mol #"NH"_4^"+"# in 100 mL solution.
#["NH"_4^"+"] = "0.05 mol/L"#
#color(white)(mmmmmmll)"NH"_4^"+" + "H"_2"O" ⇌ "NH"_3 +"H"_3"O"^"+"#
#"I/mol·L"^"-1": color(white)(mll)0.05color(white)(mmmmmmll)0color(white)(mmm)0#
#"C/mol·L"^"-1": color(white)(mll)"-"xcolor(white)(mmmmmmll)"+"xcolor(white)(mm)"+"x#
#"E/mol·L"^"-1": color(white)(m)"0.05-"xcolor(white)(mmmmmm)xcolor(white)(mmm)x#
#K_text(a) = K_text(w)/K_text(b) = (1.00 × 10^"-14")/(1.76 × 10^"-5") = 5.68 × 10^"-10"#
#K_text(a) = (["NH"_3]["H"_3"O"^"+"])/(["NH"_4^"+"]) = x^2/(0.05-x) = 5.68 × 10^"-10"#
#x^2 = 0.05 × 5.68 × 10^"-10" = 2.8 × 10^"-11"#
#x = 5.3 × 10^"-6"#
#["H"_3"O"^"+"] = xcolor(white)(l) "mol/L" = 5.3 × 10^"-6" color(white)(l)"mol/L"#
#"pH" = "-log"(5.3 × 10^"-6") = 5.27#
(iv) pH at 60 mL #"HCl"#
We have added 0.006 mol of #"HCl"#, and 0.005 mol have been neutralized by the #"NH"_3#.
Thus, we have 0.001 mol of #"HCl"# in 110 mL of solution.
#["H"_3"O"^"+"] = "0.001 mol"/"0.110 L" = 9.1 × 10^"-3" color(white)(l)"mol/L"#
#"pH = "=log"(9.1 × 10^"-3") = 2.04#
We leave it as an exercise for the student to calculate the pH at other volumes of #"HCl"#.
d- Titration table
Here's a table I created in Excel (not all the volumes are yours).
#ulbb("V/mL"color(white)(m)"pH")#
#color(white)(m)0.0color(white)(ml)11.12#
#color(white)(ll)12.5color(white)(mll) 9.72#
#color(white)(ll)25.0color(white)(mll) 9.24#
#color(white)(ll)45.0color(white)(mll) 8.29#
#color(white)(ll)50.0color(white)(mll) 5.27#
#color(white)(ll)55.0color(white)(mll) 2.32#
#color(white)(ll)75.0color(white)(mll) 1.70#
#color(white)(l)100.0color(white)(ml)1.48#
e- Construct the titration curve
Here's what I got.
f- Identify the end point on the curve
It is approximately at the position of the red dot.
