Very briefly, explain the differences between a buffered solution and an un-buff
ID: 1048619 • Letter: V
Question
Very briefly, explain the differences between a buffered solution and an un-buffered solution in terms of changes to the pH when small amounts of 3 M HCI solution is added? Use the periodic table in the room to calculate the molar mass of oxalic acid dihydrate H_2C_2O_4. 2H_2O (give your answer to two decimal places) How many grams of solid oxalic acid dihydrate are required to be dissolved in water to prepare 250.0 mL of 0.05000 M acid solution? How many moles of oxalic acid are contained in 12.0 mL of this solution? Write the balanced total molecular equation, including the state of each species, for the complete neutralization of aqueous oxalic acid by aqueous sodium hydroxide: Calculate the volume in mL, of the NaOH solution (assume 0.100 M) that is required to completely neutralize 12.0 mL of the 0.05000 M oxalic acid solution.Explanation / Answer
molar mass of oxlalic acid dihydrate(C2H2O4 2H2O) = 12*2 +1*6 +16*6 = 126g/mole
molarity = weight of ssolute *1000/gram molar mass * volume of solution in ml
0.05 = x*1000/126*250
x = 0.05*126*250/1000 = 1.575g
molarity = no of moles*1000/volume in ml
0.05 = no of moles*1000/12
no of moles = 0.05*12/1000 = 0.0006moles
2NaOH + C2H2O4 --------> Na2C2O4 + 2H2O
2 moles 1mole
no of moles of oxalic acid = molarity* volume in L
= 0.05*0.012 = 0.0006 moles
1 mole of C2H2O4 react with 2 moles of NaOH
0.0006 moles of C2H2O4 react with = 2*0.0006 = 0.0012 moles of NaOH
molarity = no of moles*1000/volume in ml
0.1 = 0.0012*1000/volume in ml
volume in ml = 0.0012*1000/0.1 = 12ml >>>> answer
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