Mastering Titration Calculations: A Chemical Analysis Guide
Expert reviewed • 22 November 2024 • 5 minute read
VIDEO
Understanding the Rough Titration
A rough titration serves as the initial trial in a titration experiment. Its primary purpose is to estimate the approximate volume needed to reach the endpoint, indicated by the indicator's color change. This preliminary trial helps analysts perform more precise measurements in subsequent trials. Important note: The rough titration result should never be included in final calculations.
Essential Calculations: Step-by-Step Examples
Example 1: Nitric Acid Concentration Analysis
Given conditions:
Standard solution: 25.0 mL of 0.050 mol/L sodium carbonate
Average titre (excluding rough titration): 22.38 mL
The calculation process follows these steps:
Balanced equation:
2 H N O 3 ( a q ) + N a 2 C O 3 ( a q ) → 2 N a N O 3 ( a q ) + H 2 O ( l ) + C O 2 ( g ) 2HNO_3(aq) + Na_2CO_3(aq) \rightarrow 2NaNO_3(aq) + H_2O(l) + CO_2(g) 2 H N O 3 ( a q ) + N a 2 C O 3 ( a q ) → 2 N a N O 3 ( a q ) + H 2 O ( l ) + C O 2 ( g )
Calculate moles of standard solution:
n ( N a 2 C O 3 ) = 0.050 × 0.025 = 0.00125 mol n(Na_2CO_3) = 0.050 \times 0.025 = 0.00125 \text{ mol} n ( N a 2 C O 3 ) = 0.050 × 0.025 = 0.00125 mol
Apply stoichiometric ratio (2:1):
n ( H N O 3 ) = 0.00125 × 2 = 0.00250 mol n(HNO_3) = 0.00125 \times 2 = 0.00250 \text{ mol} n ( H N O 3 ) = 0.00125 × 2 = 0.00250 mol
Calculate final concentration:
c ( H N O 3 ) = 0.00250 0.02238 = 0.11 mol L − 1 c(HNO_3) = \frac{0.00250}{0.02238} = 0.11 \text{ mol L}^{-1} c ( H N O 3 ) = 0.02238 0.00250 = 0.11 mol L − 1
Example 2: Hydrofluoric Acid Analysis
Using 30.0 mL of hydrofluoric acid against 0.0350 mol/L sodium carbonate:
Average titre calculation (excluding rough):
Average titre = 26.75 + 26.70 + 26.80 3 = 26.75 mL \text{Average titre} = \frac{26.75 + 26.70 + 26.80}{3} = 26.75 \text{ mL} Average titre = 3 26.75 + 26.70 + 26.80 = 26.75 mL
Balanced equation:
2 H F ( a q ) + N a 2 C O 3 ( a q ) → 2 N a F ( a q ) + H 2 O ( l ) + C O 2 ( g ) 2HF(aq) + Na_2CO_3(aq) \rightarrow 2NaF(aq) + H_2O(l) + CO_2(g) 2 H F ( a q ) + N a 2 C O 3 ( a q ) → 2 N a F ( a q ) + H 2 O ( l ) + C O 2 ( g )
Standard solution moles:
n ( N a 2 C O 3 ) = 0.0350 × 0.02675 = 9.36 × 1 0 − 4 mol n(Na_2CO_3) = 0.0350 \times 0.02675 = 9.36 \times 10^{-4} \text{ mol} n ( N a 2 C O 3 ) = 0.0350 × 0.02675 = 9.36 × 1 0 − 4 mol
Apply stoichiometry (2:1):
n ( H F ) = 9.36 × 1 0 − 4 × 2 = 1.87 × 1 0 − 3 mol n(HF) = 9.36 \times 10^{-4} \times 2 = 1.87 \times 10^{-3} \text{ mol} n ( H F ) = 9.36 × 1 0 − 4 × 2 = 1.87 × 1 0 − 3 mol
Final concentration:
c ( H F ) = 1.87 × 1 0 − 3 0.030 = 0.0624 mol L − 1 c(HF) = \frac{1.87 \times 10^{-3}}{0.030} = 0.0624 \text{ mol L}^{-1} c ( H F ) = 0.030 1.87 × 1 0 − 3 = 0.0624 mol L − 1
Example 3: Dilution Analysis with Sulfuric Acid
Given conditions:
Original solution: 1.0 L sulfuric acid
Dilution: 10.0 mL to 100.0 mL
Titration volume: 25.0 mL aliquots
Standard solution: 0.100 mol/L NaHCO₃
Average titre: 33.40 mL
Balanced equation:
H 2 S O 4 ( a q ) + 2 N a H C O 3 ( a q ) → N a 2 S O 4 ( a q ) + 2 H 2 O ( l ) + 2 C O 2 ( g ) H_2SO_4(aq) + 2NaHCO_3(aq) \rightarrow Na_2SO_4(aq) + 2H_2O(l) + 2CO_2(g) H 2 S O 4 ( a q ) + 2 N a H C O 3 ( a q ) → N a 2 S O 4 ( a q ) + 2 H 2 O ( l ) + 2 C O 2 ( g )
Standard solution moles:
n ( N a H C O 3 ) = 0.100 × 0.03340 = 0.003340 mol n(NaHCO_3) = 0.100 \times 0.03340 = 0.003340 \text{ mol} n ( N a H C O 3 ) = 0.100 × 0.03340 = 0.003340 mol
Apply stoichiometry:
n ( H 2 S O 4 ) = 0.003340 2 = 0.00167 mol n(H_2SO_4) = \frac{0.003340}{2} = 0.00167 \text{ mol} n ( H 2 S O 4 ) = 2 0.003340 = 0.00167 mol
Calculate diluted concentration:
c ( H 2 S O 4 ) diluted = 0.00167 0.025 = 0.0668 mol L − 1 c(H_2SO_4)_{\text{diluted}} = \frac{0.00167}{0.025} = 0.0668 \text{ mol L}^{-1} c ( H 2 S O 4 ) diluted = 0.025 0.00167 = 0.0668 mol L − 1
Account for dilution factor (×10):
c ( H 2 S O 4 ) original = 0.0668 × 10 = 0.668 mol L − 1 c(H_2SO_4)_{\text{original}} = 0.0668 \times 10 = 0.668 \text{ mol L}^{-1} c ( H 2 S O 4 ) original = 0.0668 × 10 = 0.668 mol L − 1
Example 4: Analysis of Citric Acid in Lemon Juice
Given conditions:
Sample: 25.0 mL lemon juice
Standard: 0.567 mol/L NaOH
Mean titre: 28.50 mL
Citric acid molar mass: 192.12 g/mol
Account for triprotic nature (1:3 ratio)
Calculate standard solution moles:
n ( N a O H ) = 0.567 × 0.02850 = 0.0162 mol n(NaOH) = 0.567 \times 0.02850 = 0.0162 \text{ mol} n ( N a O H ) = 0.567 × 0.02850 = 0.0162 mol
Apply stoichiometry:
n ( CA ) = 0.0162 3 = 0.00539 mol n(\text{CA}) = \frac{0.0162}{3} = 0.00539 \text{ mol} n ( CA ) = 3 0.0162 = 0.00539 mol
Calculate mass:
m ( CA ) = 0.00539 × 192.12 = 1.03 g m(\text{CA}) = 0.00539 \times 192.12 = 1.03 \text{ g} m ( CA ) = 0.00539 × 192.12 = 1.03 g
Final concentration:
c ( CA ) = 1.03 0.025 = 41.4 g L − 1 c(\text{CA}) = \frac{1.03}{0.025} = 41.4 \text{ g L}^{-1} c ( CA ) = 0.025 1.03 = 41.4 g L − 1