Standard Solutions and Titration

[Music] thank you the topic of this film is standard Solutions and titration sometimes a solution can be prepared by accurately weighing out the solute and then adding a precisely measured quantity of solvent many compounds are deliquescent however and cannot be weighed accurately sodium hydroxide is such a compound in this case a solution of approximately known strength is prepared and its strength is determined by titrating it against a solution of known strength or a solution containing an accurately measured quantity of solute in volumetric analysis as illustrated by these experiments the normality system of measuring concentrations is used recall that the normality of a solution is the number of gram equivalent weights of solute in a liter of solution our first problem is the determination of the normality of a sodium hydroxide solution we will add about six grams of sodium hydroxide pellets to about 300 milliliters of water and swirl to dissolve this will give us a solution about point four or 0.5 normal to standardize this solution that means to determine its strength we will use potassium acid phthalate or potassium biphthalate the formula for this compound is on the board and you can see that it's rather complicated it is however the half potassium salt of the diabasic acid phthalic acid the formula for this compound may be more conveniently written as k h c8h4o4 where this hydrogen corresponds to this hydrogen in the formula the molecular weight of this compound is 204.22 and since it contains only one acid hydrogen if the equivalent weight and its molecular weight are the same having a compound with a Heim equivalent weight is an advantage since rather large samples of it must be weighed out and this reduces weighing errors the equation for the reaction which we are about to carry out is potassium hydrogen phthalate plus sodium hydroxide produces water and potassium sodium family this is a straightforward neutralization reaction in this Beaker we have an accurately weighed sample of potassium biphthalate this sample weighs 3.0068 grams we will dissolve this material in hot distilled water while the acid is dissolving we will fill the buret with our sodium hydroxide solution the buret has previously been rinsed out with a sample of the solution we will now adjust the level of the liquid in the buret so that the bottom of the meniscus just touches the zero line on the buret we will now place the acid solution beneath the sodium hydroxide buret and phenolphthalein solution as an indicator and run in the sodium hydroxide solution until one drop produces a permanent pink color the first portion of the sodium hydroxide can be added rapidly it's obvious that a large excessive acid exists in the beaker perhaps you can begin to see a pink tinge developing in the area in the solution around where the sodium hydroxide is coming in as we approach the end point sodium hydroxide must be added more and more cautiously we're now reaching a point where just a few more drops of sodium hydroxide will cause the solution to assume a permanent pink color this is the end point of the titration subgurat now reads 26.24 milliliters from the data which we have assemble as a result of doing this experiment we can now proceed to calculate the normality of our sodium hydroxide solution and in so doing we will make use of the basic equation used in almost all acid-base titration experiments and that is that the equivalence of Base used at the endpoint or at neutrality are exactly equal to the equivalence of acid used this equation will occur several times in the course of this film now the equivalence of bays are sodium hydroxide solution can be calculated if we know the normality of the base which is the number of equivalents per liter and the volume of the base in liters the equivalence of acid present in the reaction can be calculated because we know the actual weight in grams of acid taken and we know the gram equivalent weight of the acid and this quotient will represent the equivalence of acid used now this equation contains four terms as you can see normality of the base volume of the base grams of acid and gram equivalent weight of the acid of these four terms we know three because we've measured the volume of the base that we used it was 26.24 milliliters or 0.02624 liters we know the weight of the acid it was 3.0068 grams and the equivalent weight of phthalic acid or potassium by phthalate is 204.22 grams the normality of the base is the value that we're trying to calculate here and it's the remaining term in this equation which we don't know so simply substituting in here and rearranging giving it this this fraction for the normality of the base and carrying out the indicated computations here yields the answer to this particular problem the normality of the base 5.5611 and we will now proceed to use this value in further calculations uh and problems in actual lab practice a second sample of acid would be used and a checking result obtained for the normality of the base but we will omit that step here we will now use our standardized sodium hydroxide to determine the acid strength of vinegar which contains acetic acid we will now add the vinegar to the burap on the left of the screen this buret has previously been rinsed out with things we will then adjust the level of the vinegar in the buret to zero as before next we place the clean beaker beneath the buret and measure out approximately 25 species of any to the 25 milliliters of vinegar in the beaker we now add some distilled water to bring the volume to convenient size we transfer the beaker with the vinegar to the sodium hydroxide if you're at which has previously been filled and adjusted to zero phenolphthaleine solution is added anatoleium hydroxide added to the vinegar quite rapidly after the end point has been about reached we can return to the vinegar buret and at this time one drop of vinegar should change the color from Pink back to colorless one drop pink color disappears but now we can return to the sodium hydroxide we're at carefully add one drop sodium hydroxide one more and re-establish the pink color this is the end point of this titration we will now read both burette the vinegar buret reads 25.23 milliliters while the sodium hydroxide you're at reads 30.62 milliliters we can calculate the normality of the vinegar using the formula the volume of the acid in liters times the normality of the acid the acid in each case being the vinegar equals the volume of the base times the normality of the base the volume again being in meters this formula boils down to saying that equivalence of acid used equals equivalent to base use since we saw before that the volume of the acid in liters times the normality which is the number of equivalents per liter equals the number of equivalents of acid used now in this formula we have four terms and as a result of our experiment in previous calculations we know three of them the volume of the acid was just measured and found to be 25.23 milliliters or 0.02523 liters the normality of the acid [Applause] is the value that we are attempting to find the volume of the base we found to be 30.62 milliliters or 0.03062 liters and the normality of the base we calculated previously to be 0.5611 solving this equation for the normality of the acid we find the normality of the acid is equal to 0.03062 times 0.5611 divided by 0.02523 [Music] [Applause] and carrying out the necessary computations we find that the normality of the acid is 0.6 809 this then is the strength of the vinegar now having calculated the normality of the acetic acid in our vinegar we can now proceed to calculate the percent acetic acid by weight in this particular sample acetic acid as you know has the formula hc2h3o2 but only one of the hydrogens is an acid hydrogen and therefore the molecular weight of acetic acid is identical with its equivalent weight and both are equal to 60. in the previous experiment we found this normality of this particular sample of vinegar was .6809 this means of course that .6809 equivalence of acetic acid would be present in one liter of the spinninger well knowing the number of equivalents present in a liter and the weight of one equivalent it's a simple matter to calculate the actual weight of acetic acid present in a liter of this vinegar the weight of the acid present per liter being the normality the number of equivalents per liter multiplied by the weight of one equivalent or 60. and carrying out this multiplication we find that 40.9 grams of acetic acid would be present in one liter of the vinegar now at this point we must make a modest assumption we need to assume something about the density of this vinegar solution because we haven't measured it so we'll assume that the density of this vinegar is the same as that of pure water which is not an unreasonable assumption therefore one liter of this vinegar would weigh one thousand grams and we can calculate the approximate percent uh by weight of acetic acid in this vinegar by taking the weight of acetic acid present per liter dividing it by the weight of one liter of vinegar which we've assumed to be a thousand grams and multiplying by a hundred to get percent and when we do this we find that the at least the approximate percentage of vinegar in this sample of acetic acid in the sample vinegar is 4.09 percent now this degree is rather well with manufacturer's claims for this particular sample of vinegar because on the label this vinegar is said to contain four percent acetic acid by weight and our calculations and experiments have borne this claim out assume that your instructor has given you a solid acid and told you that it might be oxalic or malonic or succinic acid these acids have molecular weights of 90. 104 and 118 respectively and since each contain two acid hydrogens the equivalent weights are in each case one half of the molecular weight or 45 52 or 59 respectively these acids are similar in properties but can be differentiated because of their difference in equivalent weight in this Beaker we have a carefully weighed sample of the unknown acid this sample weighs [Music] 0.6152 grams we will dissolve this sample in distilled water then add phenolphthalein and titrate as before sodium hydroxide buret has previously been filled and adjusted to the zero level at the end point we have used 21.12 milliliters of sodium hydroxide solution we can now calculate the equivalent weight of our unknown acid as follows as before the number of equivalents of acids used at the end point equal the number of equivalents of Base use the number of equivalents of acid is given by the weight of the acid divided by the gram equivalent weight of the acid and as before the number of equivalents a base used is equal to the volume of the base in liters times the normality of the base the number of equivalents per liter I love these four terms we know three the equation can be rearranged solving for the gram equivalent weight of the acid to give the grams of the acid over the volume of the base times the normality of the base our sample of acid weight 0.6152 grams the volume of the base was 21.12 milliliters or 0.02112 liters and the normality of the base we have previously determined to be 0.5611 solving this fraction gives the result then the gram equivalent weight of the acid as being equal to 51.9 and this obviously means that our unknown acid was malonic acid in this film we have seen how sodium hydroxide Solutions are standardized and how a standard solution may be used to determine the normality of an acid solution in this case vinegar and also how the standard solution of Base may be used to determine the equivalent weight of an unknown acid foreign [Applause]