1) You need to prepare an acetate buffer of pH 5.85 from a 0.756 M acetic acid s
ID: 474452 • Letter: 1
Question
1) You need to prepare an acetate buffer of pH 5.85 from a 0.756 M acetic acid solution and a 2.18 M KOH solution. If you have 975 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 5.85? The pKa of acetic acid is 4.76.
2) What are the concentrations of acetic acid (pKa = 4.76) and acetate in a buffer solution of 0.20 M at pH 4.9? Enter your answer to two significant figures. (For example, 0.0090 and 0.0097 each have two significant figures, whereas 0.009 has only one significant figure.)
3) What is the pH of 40.0 mL of a solution that is 0.15 M in CN– and 0.25 M in HCN? For HCN, use Ka = 4.9 × 10–9.
4) You have a buffer solution composed of 3.00 mol of acid and 5.75 mol of the conjugate base. If the pKa of the acid is 2.80, what is the pH of the buffer?
You need to prepare an acetate buffer of pH 5.85 from a 0.756 M acetic acid solution and a 2.18 M KOH solution. If you have 975 mL of the acetic acid solution, how many milliliters of the KOH solution do you need to add to make a bufer of pH 5.85? The pKa of acetic acid is 4.76. Number ml,Explanation / Answer
You have 975 mL x 0.756 M = 0.7371moles acetic acid .
........CH3COOH + KOH ==> CH3COONa + H2O
initial.0.7371 ..0.......0......0
add............ x................
change..-x.....-x......+x.....+x
equil..0.7371 -x....0........x.....x
pH = pKa + log [(base)/(acid)]
5.85 = 4.76 + log(x/0.7371-x)
1.09= log(x/0.7371-x)
(x/0.7371-x) =12.30
x= 9.066 -12.30 x
13.30 x= 9.066
x=0.681 moles
this is the number of mole sof KOH
Molarity = number of mole s / volume in L
Volume in L = Number of moles / molarity
= 0.681 moles /2.18 mole/L
= 0.2835 L
= 283.5 ml
Total volume = 283.5+ 975 ml= 1285.5 L
Cross check
pH = pKa + log [(base)/(acid)]
= 4.76 + log(0.681 moles/1285.5 L /0.7371-0.681 /1285.5 L)
=4.76 + log(0.681 /0..0561
=4.76+log 12.14
=4.76+1.084
=5.844
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