# Question f77cd

Sep 4, 2017

The concentration of the phosphoric acid is 1.23 × 10^"-3" color(white)(l)"mol·dm"^"-3".

Step 1. Write the balanced equation

${M}_{\textrm{r}} : \textcolor{w h i t e}{m l l} 310.18 \textcolor{w h i t e}{m m m m m m m m m m m m m m m m} 98.00$
$\textcolor{w h i t e}{m m m} {\text{Ca"_3("PO"_4)_2 + "3HNO"_3 → "6Ca(NO"_3)_2 + "2H"_3"PO}}_{4}$

Step 2. Calculate the moles of "Ca"_3("PO"_4)_2

"Moles of Ca"_3("PO"_4)_2 = 7.26 color(red)(cancel(color(black)("g Ca"_3("PO"_4)_2))) × (1 "mol Ca"_3("PO"_4)_2)/(310.18 color(red)(cancel(color(black)("g Ca"_3("PO"_4)_2)))) = "0.023 41 mol Ca"_3("PO"_4)_2

Step 3. Calculate the moles of ${\text{H"_3"PO}}_{4}$

"Moles of H"_3"PO"_4 = "0.023 41" color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2))) × ("2 mol H"_3"PO"_4)/(1 color(red)(cancel(color(black)("mol Ca"_3("PO"_4)_2)))) = "0.046 81 mol H"_3 "PO"_4

Step 4. Calculate the molarity of the "H"_3 "PO"_4#

$\text{Molarity" = "moles"/"cubic decimetres" = "0.046 81 mol"/"38.0 dm"^3 = 1.23 × 10^"-3" color(white)(l)"mol·dm"^"-3}$