You are to prepare 100. mL of an acetate buffer of pH 5.00 using 1.0 M acetic ac
ID: 509576 • Letter: Y
Question
You are to prepare 100. mL of an acetate buffer of pH 5.00 using 1.0 M acetic acid (HC_2H_3O_2) and solid sodium acetate (NaC_2H_3O_2). To prepare for the investigation, complete the following calculations. The pKa for HC_2H_3O_2 is 4.75. (a) What ratio of [NaC_2H_3O_2] to [HC_2H_3O_2] 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 HC_2H_3O_2 would be needed to prepare 100. mL of buffer with an acid concentration of 0.10 M? mL (c) How many grams of NaC_2H_3O_2 middot 3 H_2O would be needed to prepare 100. mL of buffer if the acid concentration is 0.10 M? g (d) If you add 2.4 mL of 0.50 M NaOH to 25.0 mL of the buffer, what is the new pH?Explanation / Answer
since pH= pKa+ log [A-]/[HA], where [A-] is conjugate base of acetic acid
5=4.75+ log[ [A-]/[HA]
[A-]/[HA] = 1.78
2. Let the volume of 1M acetic acid = xml
x*1/1000 = moles of acetic acid = 0.1*100/100= 0.1*0.1
hence x= 0.1*0.1*1000= 10ml
3. moles of acid = 10*1/1000 =0.01, let x= mass of sodium acetate
moles of sodium acetate= mass/molar mass = x/82
concentrations : acetic acid =0.1M and sodium acetate= x/82/0.1= 0.122xM
since [A-]/[HA]=1.78
0.122x/0.1= 1.78
0.122x= 0.178, x= 0.178/0.122= 1.46 gm
buffer contains 1.46 gm of sodium acetate and 0.1M acetic acid
moles of acetic acid in 25ml =0.1*25/1000 =0.0025 moles of sodium acetate= 1.46/82=0.018
Acetic acid reacts with NaOH as per CH3COOH+NaOH------>CH3COONa+H2O
moles of sodium acetate= 0.5*2.4/1000 =0.0012
NaOH is the limiting reactants since its moles are 0.0012 as against 0.0025 moles of acetic acid
moles of sodium acetate formed = 0.0012, moles of acetic acid remaining =0.0025-0.0012=0.0013
concentrations : acetic acid : 0.0013*1000/25=0.052, sodium acetate= 0.0012*1000/25= 0.048
pH= 4.75+log (0.048/0.052)= 4.715
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