C consider the titration of 250 mlml of 00200 mm h2co3h2co3...

c) Consider the titration of 25.0 mL of 0.0200 M H2CO3 with 0.0250 M KOH. Calculate the pH after the addition of 30.0 mL of base. (Ka1=4.3×10−7 and Ka2=5.6×10−11.) d) Consider the titration of 25.0 mL of 0.0200 M H2CO3 with 0.0250 M KOH. Calculate the pH after the addition of 40.0 mL of base. (Ka1=4.3×10−7 and Ka2=5.6×10−11.) e) Consider the titration of 25.0 mL of 0.0200 M H2CO3 with 0.0250 M KOH. Calculate the pH after the addition of 50.0 mL of base. (Ka1=4.3×10−7 and Ka2=5.6×10−11.)

Transcript

00:01 This is a weak acid, strong base titration calculation where we have various volumes of strong base that has been added, and we are asked to calculate the ph after each volume of strong base has been added.

00:18 Before any strong base has been added, 0 .0 milliliters, all we have in solution is a 0 .1 molar weak acid, so we can solve for the ph using the k -a expression rearranged so that the hydronium concentration is equal to the square root of k -a given to us at 6 .2 times 10 to the negative 10, multiplied by the weak acid concentration, which is 0 .1, and we get 7 .87 times 10 to the negative 6 molar.

00:56 Ph is then simply the negative log of this value, or 5 .10.

01:04 Next, we need to calculate the ph at 50 milliliters.

01:08 To make sense out of 50 milliliters, we need to first determine that the equivalence point is 100 milliliters.

01:15 We know this because the stoichiometry is one -to -one.

01:19 We are starting with 100 milliliters of 0 .1 molar weak acid, and we are adding 0 .1 molar of strong base.

01:27 Therefore, the volume of strong base required will be equal to the volume of weak acid we start with because their concentrations are the same, namely 0 .1.

01:38 So 50 milliliters is halfway to the equivalence point.

01:42 Something unique about halfway to the equivalence point of a weak acid strong -based titration is that ph equals pca, or the negative log of the ka value, which is 9 .21.

01:55 Now we calculate the ph after 75 milliliters has been added.

02:00 We still have some weak acid left and every mole of strong base has been converted into a mole of weak base.

02:09 So we have weak acid and weak base.

02:11 It's a buffer solution.

02:13 So ph will be equal to pca plus the log of the moles of base, which will be equal to the moles of strong base added.

02:22 The moles of strong base will be calculated by taking the volume of, sorry, the moles of weak base will be equal to the moles of strong base.

02:33 The moles of strong base is the volume of strong base added.

02:36 0 .075 multiplied, which is liters, multiplied by the concentration, 0 .1.

02:43 The moles of weak acid that will be left over will be the moles of weak acid we start with, 100 milliliters, which is 0 .1 liters at 0 .1 molar concentration, minus the moles of strong base added...

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