A solution made with 0.134 moles of a triprotic acid (HA, K1=6.54x10-2; K2=6.12x10-5; Kas 7.40108) dissolved...

  

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A solution made with 0.134 moles of a triprotic acid (HA, K1=6.54x10-2; K2=6.12x10-5; Kas 7.40×108) dissolved in 162 mL of solution, was titrated with 0.807 M NaOH. Answer the following question about the pH of the solution, during this titration experiment: (Points A-G correspond to initial, half-way and equivalence points throughout the titration.) 3) What volume of 0.807 M NaOH solution do you need to add, to get to the final equivalence point of this titration? Volume of base (mL) = number (rtol=0.03, atol=1e-08) 4) What is the pH at equivalent point G? = number (rtol=0.03, atol=1e-08) pH A H₂A/H₂A™ B H₂A H₂A-/HA²- D CH₂A V1 HA²- E HA²- 2-//43 F V2 Vol. NaOH(aq) added GA³- V3 A solution made with 0.134 moles of a triprotic acid (HA, K1=6.54x10-2; K2=6.12x10-5; Kas 7.40×108) dissolved in 162 mL of solution, was titrated with 0.807 M NaOH. Answer the following question about the pH of the solution, during this titration experiment: (Points A-G correspond to initial, half-way and equivalence points throughout the titration.) 3) What volume of 0.807 M NaOH solution do you need to add, to get to the final equivalence point of this titration? Volume of base (mL) = number (rtol=0.03, atol=1e-08) 4) What is the pH at equivalent point G? = number (rtol=0.03, atol=1e-08) pH A H₂A/H₂A™ B H₂A H₂A-/HA²- D CH₂A V1 HA²- E HA²- 2-//43 F V2 Vol. NaOH(aq) added GA³- V3

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Given For Triprotic Acid Ka1654102 Ka2612105 Ka375108 noofmolesofAcid0134 Volumeofacid162... View the full answer