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- a) The molar mass of the acid is 178.54 mol/lit.
- b) Ka = 9.48 x 10-5
- Explanation:
- a) let the molar mass of the acid is M.
- 1.291 g sample = 1.291/M mole.
- 1.291/M mole dissolved in 100 mL.
- Initially, [acid] = 12.91/M molar
- At equivalence point, V1S1 = V2S2
- 60 mL x 12.91/M = 47.14 mL x 0.09203
- M = (60 x 12.91)/(47.14 x 0.09203)
- M = 178.54
- The molar mass of the acid is 178.54 mol/lit.
- b) Initially, [acid] = 12.91/178.54 = 0.07231 M
- when 27.01 mL of base is added,
- V1S1 = V2S2
- V1 x 0.07231 = 27.01 mL x 0.09203
- V1 = 34.38 mL acid will be neutralized.
- Now, [salt] = 27.01 x 0.09203/ (27.01+34.38) = 0.04049 M
- [acid] = (60-34.38) x 0.07231/(27.01+34.38) = 0.03018 M
- Applying Henderson equation when 27.01 mL of base is added,
- pH = pKa + log{[salt]/[acid]}
- pKa = pH - log{[salt]/[acid]} = 4.15 - log (0.04049 /0.03018)
- pKa = 4.02237
- Ka = 9.48 x 10-5
- Comment Unhelpful Flag Answer
- Answered 4 hours ago by:
- Math_expert
- Molar mass of the acid is 298.14 g/mol
- pKa of the acid is 4.0225
- Explanation:
- Moles of base consumed at equivalent point = 0.09203 mol/L * 0.04714L= 0.004338 mol
- Moles of the acid in 60 ml = 0.004338 mol
- Concentration of the acid = 0.004338/60*1000 = 0.0723 mol/L
- 0.0723 mol/L is the concentration of the dilute acid.
- Concentration of the original acid = 0.0723*60 /100 = 0.0433 M
- Moles of the acid in 100 ml = 0.0433/1000*100 = 0.00433 mol
- Molar mass of the acid = mass of the acid/moles of the acid in 100 ml
- = 1.291g/0.00433 = 298.14 g/mol
- b)
- At a pH of 4.15, moles of NaOH added = 0.09203*27.00/1000 = 0.002485 moles
- 1 mole of the base react with the acid to produce 1 mole of the salt
- Moles of the salt firmed = 0.002485 moles
- Moles of unused acid = 0.004338 - 0.002485 = 0.001853 moles
- pKa = pH - log [salt] / [unused acid]
- = 4.15 - log (0.002485/0.001853)
- = 4.15 - 0.1275
- = 4.0225
- Comment Unhelpful Flag Answer
- Answered 4 hours ago by:
- livink261
- a) The molar mass of the acid = 178.56 g/mol
- b)The pKa of the acid = 4.02
- Explanation:
- a) The molar mass of the acid
- Calculate the number of moles of NaOH at equivalence point
- The equivalence point was reached with the addition of 47.14 mL of with 0.09203-M NaOH
- Number of moles of NaOH at equivalence point = C * V(liter)
- = 0.09203-M *(47.14 mL* 1L/1000ml)
- = 0.0043382942 moles
- At the equivalence point
- The moles of unknown monoprotic acid = the moles of base
- So, moles of acid = 0.0043382942 moles
- 60.0 mL of acid solution was titrated with NaOH.
- The concentration of acid = Number of moles /volume of acid solution
- = 0.0043382942 moles/(60.0 mL* 1L/1000ml)
- = 0.07230 M
- The concentration of unknown monoprotic acid = 0.07230 M
- A 1.291-g sample of a solid, weak, monoprotic acid is used to make100.0 mL of solution.
- So, concentration of solution in g/L = 1.291-g //(100.0 mL* 1L/1000ml) = 12.91 g/L
- Now calculate the molar mass of the acid
- The molar mass of the acid = 12.91g/L / 0.07230 mol/L
- = 178.56 g/mol
- b)The pKa of the acid
- Let HA is weak, monoprotic acid, the reaction of HA with base is given below
- HA + OH- ----> H2O + A-
- pH = 4.15(given)
- Number of moles of NaOH used at pH 4.15 = C * V (liter)
- = 0.09203-M *(27.01 mL* 1L/1000ml)
- = 0.0024857303 moles
- 0.0024857 moles of NaOH neutralizes 0.0024857 moles monoprotic acid to A-
- Remaining moles of weak, monoprotic = 0.0043382942 moles - 0.0024857303 moles
- = 0.0018525639 moles
- Total volume of solution = 60.0 ml + 27.01 ml = 87.01 ml
- [A-] = 0.0024857303 moles/(87.01 mL* 1L/1000ml) = 0.02857 M
- [HA] = 0.0018525639 moles/(87.01 mL* 1L/1000ml) = 0.02129 M
- By using Henderson-Hassel Balch equation, we can calculate pKa of weak, monoprotic acid.
- pH = pKa + log [A-] /[ HA]
- 4.15 = pKa + log (0.02857/0.02129)
- 4.15 = pKa + 0.13
- pKa = 4.15 - 0.13 = 4.02
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