Acid-base Titration Calculations Key Concepts acid-base reactions involve a proton transfer the acid donates a proton to the base acid-base reactions are also known as neutralisation reactions acid + base -----> salt + water acid A + base B -----> conjugate acid of base B + conjugate base of acid A
(Lowry-Brönsted theory) H+ + OH- ------> H 2O is the most general neutralisation reaction Equivalence point is the point at which the moles of H + is equal to the moles of OH -
An
indicator
is used to show the equivalence point during a
titration A titration involves the progressive addition of one reactant from a burette (usually the acid), to a known volume of the other reactant in a conical flask (usually the base) Calculations a. Write the balanced chemical equation for the reaction b. Extract all the relevant information from the question c. Check that data for consistency, for example, concentrations are usually given in M or mol L-1 but volumes are often given in mL. You will need to convert the mL to L for consistency. The easiest way to do this is to multi ply the volume in mL x 10 -3 d. Calculate the moles of reactant (n) for which you have both the volume(V) and concentration(M) : n = M x V e. From the balanced chemical equation find the mole ratio known reactant : unknown reactant f. Use the mole ratio to calculate the moles of the unknown reactant g. From the volume(V) of unknown reactant and its previously calculated moles(n), calculate its concentration(M): concentration(M): M = n ÷ V Examples 1.
30 mL of 0.10M NaOH neutralised 25.0mL of hydrochloric acid. Determine the concentration of the acid a. Write the balanced chemical equation for the reaction NaOH(aq) + HCl(aq) -----> NaCl(aq) + H 2O(l) b. Extract the relevant information from the question: NaOH V = 30mL , M = 0.10M HCl V = 25.0mL, M = ? c. Check the data for consistency NaOH V = 30 x 10-3L , M = 0.10M HCl V = 25.0 x 10-3L, M = ?
d. Calculate moles NaOH n(NaOH) = M x V = 0.10 x 30 x 10 -3 = 3 x 10 -3 moles e. From the balanced chemical equation find the mole ratio NaOH:HCl 1:1 f. Find moles HCl NaOH: HCl is 1:1 So n(NaOH) = n(HCl) = 3 x 10 -3 moles at the equivalence point g. Calculate concentration of HCl: M = n ÷ V n = 3 x 10-3 mol, V = 25.0 x 10-3L -3 M(HCl) = 3 x 10 ÷ 25.0 x 10 -3 = 0.12M or 0.12 mol L -1 2. 50mL of 0.2mol L-1 NaOH neutralised 20mL of sulfuric acid. Determine the concentration of the acid a. Write the balanced chemical equation for the reaction 2NaOH(aq) + H 2SO4(aq) -----> Na 2SO4(aq) + 2H2O(l) b. Extract the relevant information from the question: NaOH V = 50mL, M = 0.2M H2SO4 V = 20mL, M = ? c. Check the data for consistency NaOH V = 50 x 10-3L, M = 0.2M H2SO4 V = 20 x 10-3L, M = ? d. Calculate moles NaOH n(NaOH) = M x V = 0.2 x 50 x 10 -3 = 0.01 mol e. From the balanced chemical equation find the mole ratio NaOH:H2SO4 2:1 f. Find moles H2SO4 NaOH: H2SO4 is 2:1 So n(H2SO4) = ½ x n(NaOH) = ½ x 0.01 = 5 x 10 -3 moles H2SO4 at the equivalence point g. Calculate concentration of H 2SO4: M = n ÷ V n = 5 x 10-3 mol, V = 20 x 10-3L M(H2SO4) = 5 x 10-3 ÷ 20 x 10 -3 = 0.25M or 0.25 mol L -1 3. 25.0mL of 0.05M Ba(OH)2 neutralised 40.0mL of nitric acid. Determine the concentration of the acid. a. Write the balanced chemical equation for the reaction Ba(OH) 2(aq) + 2HNO3(aq) -----> Ba(NO 3)2(aq) + 2H 2O(l) b. Extract the relevant information from the question: Ba(OH)2 V = 25.0mL, M = 0.05M HNO3 V = 40.0mL, M = ? c. Check the data for consistency Ba(OH)2 V = 25.0 x 10-3L, M = 0.05M HNO3 V = 40.0 x 103 L, M = ? d. Calculate moles Ba(OH) 2 n = M x V = 0.05 x 25.0 x 10 -3 = 1.25 x 10-3 mol e. From the balanced chemical equation find the mole ratio Ba(OH) 2 : HNO 3 1:2 f. Find moles HNO3 Ba(OH) 2 : HNO 3 is 1 : 2 So n(HNO3) = 2 x n(Ba(OH) 2) = 2 x 1.25 x 10 -3 = 2.5 x 10 -3 mol g. Calculate concentration of HNO 3: M = n ÷ V n = 2.5 x 10 -3moles, V = 40.0 x 10-3L M = 2.5 x 10-3 ÷ 40.0 x 10 -3 = 0.0625M or 0.0625 mol L-1
Calculation of titration result is always based on the stoichiometry of the titration reaction. When titrating we are usually given information about volume and concentration of the t itrant solution and about volume of the titrated substance solution. Balanced reaction equation shows numbers of moles of reacting substances, thus to be able to deal with titration results we have to be able to convert between volumes, concentrations and numbers of moles. All these conversions are based on definition of molar concentration:
1 which can be rearranged to calculate number o f moles when concentration and vol ume are given: 2 (where n is number of moles, C concentration and V is volume). In most cases concentration is given in moles per liter of solution (molarity, denoted here as M), but sometimes it can be also given as normality. For details see molarity and normality lectures on ChemBuddy site.
For example let's look at the simple acid base t itration - determination of concentration o f sulfuric acid with the use of NaOH solution of known concentration C NaOH. From the titration we know volume of titrant used (V NaOH). We know that one mole of H 2SO4 reacts with exactly two moles of NaOH: 2NaOH + H2SO4
Na2SO4 + 2H2O
→
As we know that amount of substance of a given concentration in given volume of solution is C×V, and sulfuric acid reacts with sodium hydroxide in the 1:2 ratio, we can write: 3 There is only one unknown in this equation:
4
Occasionally some additional calculations are needed, as determination of concentration of titrated substance may be only a step in the assay. For example if we are asked to find a purity of the substance, we must convert concentration found to amount of substance in the original sample. These calculations are usually based on simple number of moles/mass of substance conversions, or on simple dilution calculations.
10.00 mL sample of concentrated HCl was diluted to mark in 250 mL volumetric flask. To titrate 25.0 mL sample of diluted solution 43.56 mL of 0.1023 M NaOH was needed. What was the original concentration of hydrochloric acid? We start with the reaction equation: HCl + NaOH
NaCl + H 2O
→
One mole of hydrochloric acid reacts with one mole of NaOH. For titration 0.04356 L×0.1023 M=4.456×10-3 mole of base was used, so there was 4.456 mmole of hydrochloric acid in every 25.00 mL of solution taken from the volumetric flask. Volumetric flask is 10 times larger than the samples titrated, so it contained 44.56 mmole of acid. This amount was originally found in 10.00 mL of concentrated acid - so original concentration was 44.56 mmole/10.00 mL=4.456 M (note that concentrations expressed in mmole/mL and mol/L have always the same numerical value, so to convert between them it is enough to change units).
1.456 g sample of unknown alloy was dissolved in acid and diluted to 100 mL. What was the percentage of iron in the sample, if titration of 20.00 mL aliquot of the diluted solution requires on average 33.45 mL of 0.01005 M solution of potassium permanganate? As usual, we will start with the reaction equation: 5Fe2+ + MnO4- + 8H+
5Fe3+ + Mn2+ + 4H2O
→
33.45 mL of 0.01005 M solution of potassium permanganate contains 0.03345 L×0.01005 M=3.362×10-4 moles of titrant. Iron reacts with permanganate in 5:1 ratio, so 20.00 mL aliquot contains 5×3.362×10-4=1.681×10-3 moles of iron. There were exactly five such aliquots (100.00 mL vs 20.00 mL), so amount of iron in the sample was 5×1.681×10 3
=8.404×10-3 moles, or 55.854 g/mol×8.404×10-3 mol=0.4693 g. That in turn means there was
0.4693 g/1.456 g×100%=32.23% iron in the sample.
1.354 g sample of sodium nitrate contaminated with NaCl was dissolved in small amount of water and filled to the mark in the 100 mL volumetric flask. To titrate chlorides in 10.00 mL sample 35.70 mL of 0.01021 M AgNO 3 was used. What is the percent purity of the sample? NaCl + AgNO3
AgCl + NaNO 3
→
Chlorides react with silver nitrate solution in 1:1 ratio. That means that 10.00 mL of chlorides solution contained 35.70×0.01021=0.3645 mmole of chlorides. Total sample was 10 times larger, so it contained 3.645 mmole, or 0.2130g (molar mass of NaCl being 58.33 g/mol). Sample contains 0.2130×100%/1.354=15.73% NaCl and 100%-15.73%=84.27% of sodium nitrate (assuming there are no other impurities). Such calculations are rarely really difficult, but they are always time consuming. To make them much easier you may use EBAS - Equation Balancer And Stoichiometry calculator . You may download it from ChemBuddy site and we will provide ready files with reaction equations for all assays described.
Titration
A titration is a method of analysis that will allow you to determine the precise endpoint of a reaction and therefore the precise quantity of reactant in the titration flask. A buret is used to deliver the second reactant to the flask and an indicator or pH M eter is used to detect the endpoint of the reaction.
Doing a Titration
Begin by preparing your buret, as described on the buret page. Your buret should be conditioned and filled with titrant solution. You should check for air bubbles and leaks, before proceding with the titration.
Take an initial volume reading and record it in your notebook. Before beginning a titration, you should always calculate the expected endpoint volume.
Prepare the solution to be analyzed by placing it in a clean Erlenmeyer flask or beaker. If your sample is a solid, make sure it is completely di ssoloved. Put a magnetic stirrer in the flask and add indicator.
Use the buret to deliver a stream of titrant to within a couple of mL of your expected endpoint. You will see the indicator change color when the titrant hits the solution in the flask, but the color change di sappears upon stirring.
Approach the endpoint more slowly and watch the color of your flask carefully. Use a wash bottle to rinse the sides of the flask and the tip of the buret, to be sure all titrant is mixed in the flask.
As you approach the endpoint, you may need to add a partial drop of titrant. You can do this with a rapid spin of a teflon stopcock or by partially opening the stopcock and rinsing the partial drop into the flask with a wash bottle. Ask your TA to demonstrate these techniques for you, in the lab.
Make sure you know what the endpoint should look l ike. For phenolphthalein, the endpoint is the first permanent pale pi nk. The pale pink fades in 10 to 20 minutes. If you think you might have reached the endpoint, you can record the volume reading and add another partial drop. Sometimes it is easier to tell when you have gone past the endpoint.
If the flask looks like this, you have gone too far!
When you have reached the endpoint, read the fi nal volume in the buret and record it in your notebook.
Subtract the initial volume to determine the amount of titrant delivered. Use this, the concentration of the titrant, and the stoichiometry of the titration reaction to calculate the number of moles of reactant in your analyte solution.
For more information, see the page on burets.
Titrating with a pH meter
Titration with a pH meter follows the same procedure as a titration with an indicator, except that the endpoint i s detected by a rapid change in pH, rather than the color change of an indicator.
Arrange the sample, stirrer, buret, and pH meter electrode so that you can read the pH and operate the buret with ease.
To detect the endpoint accurately, record pH vs. volume of titrant added and plot the titration curve as you titrate.
