Very briefly, explain the differences between a buffered solution an un-buffered
ID: 1048614 • Letter: V
Question
Very briefly, explain the differences between a buffered solution 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 add 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
1.A buffered solution maintains its pH despite additions of acid or base (over a range) while the pH of an unbuffered solution will respond to modest additions of acid or base.
A buffer will prevent pH changes because a buffer system is composed of a weak acid and its conjugate base....so that when a strong acid like HCl is added the H+ from the HCl will react with the base form in the buffer making the weak acid which is weak because lttle of the acid is ionized ( which means very little H+ is produced ).
2.Oxalic acid dihydrate having the formulae H2C2O4.2H20 is 2*1 +2*12 + 4*15.999 + 4*1+2*15.999 =125.994 Upto 2 decimals means 125.99
3.Molarity = w/GMW *1000/V(ml)
0.05 = w/125.99 *1000/250
w =(0.05*125.99 )/4 = 1.574gms
4. molarity = no.of moles /volume in lit
0.05 = n / (12/1000)
n = (0.05 *12) /1000 = 0.0006
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