a. A student obtains a vinegar solution having an acetic acid concentration of 0

Question

a. A student obtains a vinegar solution having an acetic acid concentration of 0.8512 M. What volume of 0.1036 M NaOH will be required to reach the equivalence point when 25.00 mL of the vinegar solution is titrated?

b. A student dilutes the vinegar solution in part a above by pipetting 10 mL of the solution into a 100 mL volumetric flask and diluting to a total volume of 100 mL. What volume of 0.1036 M NaOH will be required to reach the equivalence point when 100 mL of the diluted vinegar solution is titrated?

c. What volume of 0.1036 M NaOH will be required to reach the equivalence point when 25 mL of the diluted vinegar solution prepared in part b is titrated?  

Explanation / Answer

firstly Kd of CH3COOH = 1.8 * 10^(-5) CH3COOH= (CH3COO-) + (H+) no of millimoles of CH3COOH = 0.8512 * 25 =21.28 * 10^(-3) moles let x be the fraction of acetic acid dissociated then Kd = (21.28x)^2/21.28(1-x) = (21.28 * x^2)/(1-x) solving the quadratic in x we get x = 0.9 approx (remember 21.28 is the millimoles so take it in mole form in the above equation ) therefore amount of (H+) = 21.28 * 0.9 = 19.152 no of moles of NaOH available = 0.1036 * V therefore V in ml = 19.152/0.1036 = 184.86 ml b) 25 ml of solution contains 21.28 millimoles of acetic acid .. therefore 10ml will have 8.512 millimoles therefore we calculate the new "x" we solve the quadratic in x again to get x = 0.959 therefore no of moles of H+ = 0.959 * 8.512 * 10^(-3) = 0.1036 * V therefore V = 7.885 ml of NaOH is required .

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