Question

5. You are asked to prepare an acetate buffer solution at pH 4.6. Use the Henderson-Hasselbalch equation to calculate the ratio of sodium acetate and acetic acid required to prepare a buffer at pH 4.6. 6. If you had 500mL of 0.200M sodium acetate buffer pH4.76, how many mL of 1.00M HCI would you need to add to adjust the pH to 4.60?

Answer 1

5. pH of the buffer solution = 4.6 Ka for acetic acid = 1.8 x 10-5 pH of the buffer is determined by Henderson equation, pH = pKa+log [base] (acid] pH =-log Ka+log [CH,C00"] [CH2COOH] 4.6=-log(1.8x10“) +logo [CH,C00-] 10% [CH2COOH] 4.6=4.745+log [CH,C00"] [CH2COOH] [CH,C00"] -4.745=-0.145 [CH2COOH] [CH;col [CH,C00"] , = 10-0.145 = 0.716593 = 0.72 [CH,COOH] ratio of sodium acetate and acetic acid required = 0.72

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6.

pKa of CH3COOH = 4.745 We know that [base] pH = pKa +log [acid] [CH,C00"] 34.76= 4.745 +log $ CH2COOH] log (CH,COO 1:C0077 = 0.015 0% [CH,COOH) [CH,COO-] -100.015 = 1.036 [CH3COOH] [CH,COO-]=1.036x[CH,COOH] Given that conc. of buffer = 0.200 M That is [CH3COO-]+[CH,COOH]=0.2 1.036x[CH,COOH]+[CH3COOH]=0.2 2.036x[CH2COOH]=0.2 [CH,COOH) = 0,36 = 0.098242 [CH,C00"]=1.036x[CH2COOH]=1.036x0.098242=0.101758 [CH,COOH] = 0.098242M [CH,COO-]=0.101758M Volume of the buffer solution = 500mL = 0.50L Number of moles of [CH,COOH) present in the solution = M*V = 0.098242M X0.5 L = 0.049121 mol Number of moles of CH3COO present in the solution = M*V =0.101758Mx0.5 L = 0.049121 mol 36

Let x be the number moles of HCl to be added to the buffer to change the pH. Number of moles of CH2COOH = 0.049121 mol Number of moles of CH3COO- = 0.050879 mol CH3COO + H CH3COOH Initial 0.0508790 0.049121 Add X Change + I Equilibrium 0.050879-X 0 0.049121 + x [CH,C00"] =pH = pKa+log [CH3COOH] 0.050879 - x → 4.60 = 4.745+log 0.049121+x 0.050879-X - _0.145 10g 0,049121+x 0.050879– X – 10-0.145 = 0.7166 0.049121+x 0.050879- x=0.7166 x(0.049121+x) 0.050879-x=0.0351997 +0.7166x 0.050879-0.0351997 =0.7166x + x 1.7166x=0.015679 x=0.009134 mol number of moles of HCI=0.009134 mol Molarity of HCl = 1.00 M Volume of HCl = moles _ 0.009134 mol =0.009134 L = 9.13 mL molarity 1.00M Volume of HCl to be added = 9.13 mL =