benzoic acid (C7H6O2) is a weak acid whose ka is 6.5*10^-5. If 0.490 g of benzoi
ID: 1064178 • Letter: B
Question
benzoic acid (C7H6O2) is a weak acid whose ka is 6.5*10^-5. If 0.490 g of benzoic acid are dissolved in 250 ml of water what mass of sodium benzoate (NaC7H5O2) would need to be added to adjust the ph of of the solution to 5.20? You may assume that the added sodium benzoate doesn't significantly chenge the solution volume.
10. Benzoic acid (C7H602) is a weak acid whose Ka is 6.5 x 10 5. If 0.490 g of benzoic acid are dissolved in 250 mL of water, what mass of sodium benzoate (NaC7H502) would need to be added to adjust the pH of the solution to 5.20? You may assume that the added sodium (8 points) benzoate doesn't significantly change the solution volume.Explanation / Answer
concentration of benzoic acid in 0.490 gm in 250 ml (0.25L) = 0.49/0.25=1.96M
Ka= 6.5*10-5
pKa = 4.19
pH= pKa+ log [concentration of conjugate base/ concentration of acid]
5.2= 4.19 + log [0.196/ concentration of acid]
1.01 = log [/ concentration of conjugate base/ 0.196)]
concentration of conjugate base/0.196 = 10.23
concentration of conjugate base = 10.23*0.196=2M
moles of conjugate base = molarity* Volume (L) = 2*0.25= 0.5
Mass of c sodium benzoate = moles*molar mas
molar mass of sodium benzoate= 23+7*12+5+32 =144
Mass of sodium benzoate = 0.5*144= 72 gm
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