The balanced equation is
"Ca(OH)"_2 + "2HNO"_3 → "Ca"("NO"_3)_2 + "2H"_2"O"
or
"OH"^"-" + "H"^+ → "H"_2"O"
"Moles of OH"^"-" = 0.0200 color(red)(cancel(color(black)("L Ca(OH)"_2))) × (4.85 × 10^"-3" color(red)(cancel(color(black)("mol Ca(OH)"_2))))/(1 color(red)(cancel(color(black)("L Ca(OH)"_2)))) × ("2 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol Ca(OH)"_2)))) = 1.94 × 10^"-4" color(white)(l)"mol OH"^"-"
"Moles of H"^+ = 0.0300 color(red)(cancel(color(black)("L HNO"_3))) × (3.75 × 10^"-3" color(red)(cancel(color(black)("mol HNO"_3))))/(1 color(red)(cancel(color(black)("L HNO"_3)))) × ("1 mol H"^+)/(1 color(red)(cancel(color(black)("mol HNO"_3)))) = 1.125 × 10^"-4" color(white)(l)"mol H"^+
All of the "H"^+ will react with the "OH"^"-".
The excess moles of "OH"^"-" are
(1.94 × 10^"-4" - 1.125 × 10^"-4") "mol" = 8.15 × 10^"-5"color(white)(l) "mol"
["OH"^"-"] = "moles"/"litres" = (8.15 × 10^"-5"color(white)(l) "mol")/("0.0500 L") = "0.001 63 mol/L"
"pOH" = "-log"["OH"^"-"] = "-log(0.001 63)" = 279
"pH" = "14.00 - pH" = "14.00 - 2.79" = 11.21
This is already a long answer, so we leave it as an exercise for the student to calculate the pH for the second question.
