** answer is also not 12 g** How many grams of dipotassium succinate trihydrate
ID: 491024 • Letter: #
Question
** answer is also not 12 g**
How many grams of dipotassium succinate trihydrate (K2C4H404 3H20, MW 248.32 g/mol) must be added to 690.0 mL of a 0.0368 M succinic acid solution to produce a pH of 5.918? Succinic acid has pKa values of 4.207 (pKat) and 5.636 (pka2). Incorrect The mass you have reported is the mass of K2C4H404 3H20 required Number to completely react with the succinic acid in solution. This amount of 6.305 g succinate is enough to completely convert all the succinic acid to the intermediate ion as shown in the following reaction To achieve a buffer with a pH of 5.918, additional succinate must be added to the solution. Use the Henderson-Hasselbalch equation associated with the dissociation of the intermediate ion to the succinate ion to calculate the additional number of moles of succinate that must be added. Note that moles can be used instead of concentration in the Henderson-Hasselbalch equation because the volume of the solution is the same for both the intermediate and succinate ion. HAO log a2 CHAO The total amount of succinate is the sum of your answer and this additional amount of succinate.Explanation / Answer
Moles of succinic acid= Molarity * Volume of solution
=0.690 * 0.0368
=0.0253
uSING hANDERSON Hasselbalch equation
5.918=(pK1+pK2 ) + log [Moles/248.32 *0.690(Volume] / 0.0253
5.918=9.83 + log [m/171.34] /0.0253
-3.912=log[m/171.34]- log0.0253
-3.912=logm -log 171.34-log 0.0253
-3.912=logm-2.23+1.59
-4.552= log m
m=0.000032
Moles of succinate=0.000032
Mass in grams = Moles * Molar mass
=0.000032 *248.32
=0.0079
=0.008
=8 * 10 ^-2 grams
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.