EniG. Chemistry Assistant (http://www.ktf-split.hr/~eni/toys/chemas-e.html)
(http://www.ktf-split.hr/~eni/toys/chemas-e.html)Prepare 1000 mL of 0.1 mol/L solution of phosphoric acid by diluting 6.74 mL of 85 % H3PO4 (density 1.71 g/mL) with distilled water to a volume of 1000 mL. The dilution is therefore made by adding 993.26 mL of water. However, the volume of a liquid obtained by mixing measured volumes of diferent solutions is not always precisely the sum of the component volumes.
NOTE:
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Ortophosphoric acid
85% H3PO4
Colorless, odorless, sparkling liquid.
Molar mass = 97.9952 g/mol
CAS No. [7664-38-2]
Boiling point 158°C
Density (85%, 20°C) 1.71 g/mL
Soluble in water
pH value (20°C): <0.5
CALCULATION:
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Source (B): 85 % solution of H3PO4 (density 1.71 g/mL)
The molar concentration of a solution of a chemical species A is the number of moles of that species contained in one liter of the solution (not in one liter of the solvent).
c(A) = 0.1 mol/L
Molar concentration = (Density * Weight percent / 100) / Molecular weight
c(B) = (d(B) * w(B) / 100) / M(B)
c(B) = (1710 g/L * 85 % / 100) / 97.9952 g/mol
c(B) = 14.8323591359577 mol/L
The number of moles of solute in the diluted solution must equal the number of moles in the concentrated reagent.
V(A) * c(A) = V(B) * c(B)
V(B) = V(A) * c(A) / c(B)
V(B) = 1 L * 0.1 mol/L / 14.8323591359577 mol/L
V(B) = 6.74201582387341E-03 L
V(B) = 6.74 mL
(https://www.thevespiary.org/rhodium/Rhodium/hive/hiveboard/picproxie_docs/000481689-file_yst6.jpg)
Prepare 1000 mL of 0.1 mol/L H3PO3 by diluting 9.65 mL of 85 % H3PO3 (density 1 g/mL) with distilled water to a volume of 1000 mL. The dilution is therefore made by adding 990.35 mL of water. However, the volume of a liquid obtained by mixing measured volumes of diferent solutions is not always precisely the sum of the component volumes.
CALCULATION:
¯¯¯¯¯¯¯¯¯¯¯¯¯
Source (B): 85 % solution of H3PO3 (density 1 g/mL)
The molar concentration of a solution of a chemical species A is the number of moles of that species contained in one liter of the solution (not in one liter of the solvent).
c(A) = 0.1 mol/L
Molar concentration = (Density * Weight percent / 100) / Molecular weight
c(B) = (d(B) * w(B) / 100) / M(B)
c(B) = (1000 g/L * 85 % / 100) / 81.9958 g/mol
c(B) = 10.3663846197 mol/L
The number of moles of solute in the diluted solution must equal the number of moles in the concentrated reagent.
V(A) * c(A) = V(B) * c(B)
V(B) = V(A) * c(A) / c(B)
V(B) = 1 L * 0.1 mol/L / 10.3663846197 mol/L
V(B) = 9.64656470588235E-03 L
V(B) = 9.65 mL
