3. -9 points TCCGenChem2 6.PRE L1 003 0/3 Submissions Used My Notes You are to p
ID: 509969 • Letter: 3
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3. -9 points TCCGenChem2 6.PRE L1 003 0/3 Submissions Used My Notes You are to prepare 100. mL of an acetate buffer of pH 5.00 using 1.0 M acetic acid (HC2H302) and solid sodium acetate (Nac2H302). To prepare for the investigation, complete the following calculations. The pKa for HC2H302 is 4.75. (a) What ratio of CNac2H302] to [HC2H302] is required? (b) You will want the concentrations relatively dilute, so use an acid concentration of 0.10 M. What volume of 1.0 M HC2H302 would be needed to prepare 100. mL of buffer with an acid concentration of 0.10 M? mL (c) How many grams of NaC2H302 3 H20 would be needed to prepare 100. mL of buffer if the acid concentration is 0.10 M? (d) If you add 2.4 mL of 0.500 M NaOH to 25.0 mL of the buffer, what is the new pH? Supporting Materials Periodic Table Additional Materials aHow well can a Buffer Resist pHChange?Explanation / Answer
a) Henderson equation is
pH = pKa + log ( [A-]/[HA] )
5 = 4.75 + log([A-]/[HA])
[A-]/[HA] = 1.78
b) 1M to 0.1M is 10 time dilution
Buffer volume = 100ml
Volume of 1M acid needed = 100ml/10 = 10ml
So, 10 ml of 1M acd needed
c) [A-] = 1.78[A]
[A-] = 1.78 × 0.1 = 0.178M
No of moles of A- required for 100ml = ( 0.178/1000)×100=0.0178
Molar mass of CH3COONa= 136g
Mass of CH3COONa requird = 136 × 0.0178= 2.4208
d) 0.5M NaOH = 0.5 mol in 1000ml
No of mol NaOH in 2.4ml = 0.0012
NaOH consumed by reacting with CH3COOH
0.0012M NaOH react with 0.0012M of CH3COOH
No of mol of CH3COOH in 25ml buffer = (0.1/1000)×25 =0.0025mol
Remaining mol of CH3COOH after addition of NaOH = 0.0025-0.0012 = 0.0013
Total volume = 25ml + 2.4ml = 27.4ml
[CH3COOH] = (0.0013mol/27.4ml)×1000ml = 0.0474M
[ CH3COONa] = 0.178/diution = 0.178/1.096=0.1624M
Applying in Henderson equation
pH = 4.75 + log (0.1624/0.0474)
= 5.28
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