molar concentration of a weak acid solution: 1. molar concentration of NaOH is 0
ID: 613255 • Letter: M
Question
molar concentration of a weak acid solution: 1. molar concentration of NaOH is 0.1 M 2. volue of weak acid is 25.0 mL 3. buret reading of NaOH initial is 0.00 mL 4. buret reading of NaOH final is 36.0 mL 5. Volume of NaOH dispensed (mL) is 36.0 mL 6. moles of NaOH to stoichiometric point (mol) __________ 7. Moles of acid (mol) _________ 8. Molar concentration of acid (mol/L)_______________ Molar mass and the pKa of a solid weak acid: 1. Mass of dry, solid acid (g) 0.312 g 2. molar cocnetration of NaOH is 0.1 M 3. buret reading of NaOH initial is 0.00 mL 4. buret reading of NaOH final is 25.0 mL 5. volume of NaOH dispensed is 25.0 mL 6. Moles of NaOH to stoichiometric pt (mol) ______ 7. moles of acid (mol) ____ 8. molar mass of acid (g/mol)______ 9. volue of NaOH half way to tstoichiometric point_________ I want to check my answer it doesn't look right when I did it the calculations.Explanation / Answer
(a) When you put acetic acid (I'll abbreviate it as HAc) in water, it dissociates to a small extent to form ions: HAc + H2O H3O+ + Ac- A small amount of H3O+ and Ac- will form, and the HAc concentration will be reduced by the same amount. Molarity . . .HAc + H2O H3O+ + A- Initial . . . . .2.0 . . . . . . . . . . . . 0 . . . . .0 Change . . . -x . . . . . . . . . . . . +x . . . .+x At equil. . .2.0 - x . . . . . . . . . . +x . . . .+x Ka = 1.8 x 10^-5 = [H3O+][A-] / [HA] = x^2 / (2.0 - x) x will be small comapred to 2.0, so eliminate the -x term after 2.0. (Rule of Thumb: If M / Ka is greater than 100, eliminate the -x term. Here, 2.0 / 1.8 x 10^-5 = 110,000.). 1.8 x 10^-5 = x^2 / 2.0 3.6 x 10^-5 = x^2 0.0060 = x = [H3O+] pH = -log [H3O+] = -log (0.0060) = 2.2 ======================================… (b) Calculate moles of each species after the HAc and NaOH react. HAc + NaOH ==> H2O + NaAc Initial moles HAc = M HAc x L HAc = (2.0)(0.100) = 0.20 moles HAc Initial moles NaOH = M NaOH x L NaOH = (1.0)(0.100) = 0.10 moles NaOH During the reaction, The 0.10 moles of NaOH will completely react with 0.10 moles of the HAc, forming 0.10 moles of NaAc and leaving 0.10 moles of HAc unreacted. When the reaction is over, M HAc = moles HAc / L = 0.10 / 0.2 = 0.50 M M NaAc = moles NaAc / L = 0.10 / 0.20 = 0.50 M NaAc To calculate the pH of a buffer, use the Henderson-Hasselbalch equation: pH = pKa + log ([A-] / [HA]) where A- is the base species in the buffer (NaAc for us) and HA is the acid species in the buffer (HAc for us). pH = -log (1.8 x 10^-5) + log (0.50 / 0.50) pH = 4.7 + log 1 pH = 4.7 + 0 = 4.7 [H3O+] = 10^-pH = 10^(-4.7) = 2.0 x 10^-5 M ======================================… (c) If you add strong acid to a buffer, it will react with the base species in the buffer (for us, NaAc). HCl + NaAc ==> HAc + NaCl Initial moles HCl = M HCl x L HCl = (0.50)(0.10) = 0.050 moles HCl Initial moles HAc = M HAc x L HAc = (0.50)(0.040) = 0.020 moles HAc Initial moles NaAc = M NaAc x L NaAc = (0.50)(0.040) = 0.020 moles NaAc During the reaction, HCl will react with the NaAc and USE IT ALL UP! That is, the 0.020 moles of NaAc will react with 0.020 moles of HCl to form 0.020 moles of HAc and leave 0.030 moles of HCl unreacted. In essence, the buffer has been destroyed by adding too much HCl. The pH of the solution will be governed by the amount of strong acid HCl that is present. The presence of the weak acid HAc will have little effect on the pH. [H3O+] = moles HCl / L = 0.030 / 0.14 = 0.21 M I think this problem has a typo in it. Way too much HCl was added.Related Questions
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