# Question eddcd

Jul 3, 2016

The pH of the resulting solution is 12.21.

#### Explanation:

The balanced equation is

$\text{Ca(OH)"_2 + "2HNO"_3 → "Ca"("NO"_3)_2 + "2H"_2"O}$

or

$\text{OH"^"-" + "H"^+ → "H"_2"O}$

$\text{Moles of OH"^"-" = 0.0200 color(red)(cancel(color(black)("L Ca(OH)"_2))) × (4.85 × 10^"-3" color(red)(cancel(color(black)("mol Ca(OH)"_2))))/(1 color(red)(cancel(color(black)("L Ca(OH)"_2)))) × ("2 mol OH"^"-")/(1 color(red)(cancel(color(black)("mol Ca(OH)"_2)))) = 1.94 × 10^"-4" color(white)(l)"mol OH"^"-}$

${\text{Moles of H"^+ = 0.0300 color(red)(cancel(color(black)("L HNO"_3))) × (3.75 × 10^"-3" color(red)(cancel(color(black)("mol HNO"_3))))/(1 color(red)(cancel(color(black)("L HNO"_3)))) × ("1 mol H"^+)/(1 color(red)(cancel(color(black)("mol HNO"_3)))) = 1.125 × 10^"-4" color(white)(l)"mol H}}^{+}$

All of the ${\text{H}}^{+}$ will react with the $\text{OH"^"-}$.

The excess moles of $\text{OH"^"-}$ 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"#

$\text{pOH" = "-log"["OH"^"-"] = "-log(0.001 63)} = 279$

$\text{pH" = "14.00 - pH" = "14.00 - 2.79} = 11.21$

This is already a long answer, so we leave it as an exercise for the student to calculate the pH for the second question.