Construct a curve for the titration of 20.00 mL 0.500 N succinic acid (HO2CCH2CH2CO2H) by 0.1000 N NaOH solution. Use MS Excel for convenience. Titrant volumes: 0.00 mL, 5.00 mL, 10.00 mL, 15.00 mL, 20.00 mL, 25.00 mL, and 30.00 mL. (CH2)2(CO2H)2⇄(CH2)2(CO2H)(CO2)^- +H^+ Ka1= 6.21 x 10^-5; pKa1 =4.207(CH2)2(CO2H)(CO2)^-⇄(CH2)2(CO2)2^2-+ H^+ Ka2= 2.31 x 10^-6; pKa2 =5.636 3. Construct a curve for the titration of 20.00 mL 0.0500 N succinic acid (HO;CCH;CH,CO,H)by 0.1000 N NaOH solution. Use MS Excel® for convenience. Titrant volumes: 0.00 mL,5.00 mL, 10.00 mL, 15.00 mL, 20.00 ml, 25.00 mL, and 30.00 mL.(CH)2(CO2H)2 = (CH2)2(CO2H)(CO2) + H*(CH.)2(CO,H)(CO.) (CH2)2(CO2), + H*Kai = 6.21 x 10s; pKai = 4.207Ka2 = 2.31 x 106; pK2 = 5.636%3!%3D%3!

We can build the bend for titration of corrosive against a base by taking the pH on the y-pivot and…

Construct a curve for the titration of 20.00 mL 0.0500 N succinic acid (HO;CCH;CH;,CO,H) by 0.1000 N NaOH solution. Use MS Excel® for convenience. Titrant volumes: 0.00 mL, 5.00 mL, 10.00 mL, 15.00 mL, 20.00 ml, 25.00 ml, and 30.00 mL. (CH:)2(CO2H)2 = (CH2):(CO;H)(CO2)" + H* (CH)2(CO;H)(CO,) = (CH2);(CO2);²- + H* Ka1 = 6.21 x 10; pKs1 = 4.207 K2 = 2.31 x 10; pK2 = 5.636

Construct a curve for the titration of 20.00 mL 0.0500 N succinic acid (HO;CCH;CH;,CO,H) by 0.1000 N NaOH solution. Use MS Excel® for convenience. Titrant volumes: 0.00 mL, 5.00 mL, 10.00 mL, 15.00 mL, 20.00 ml, 25.00 ml, and 30.00 mL. (CH:)2(CO2H)2 = (CH2):(CO;H)(CO2)" + H* (CH)2(CO;H)(CO,) = (CH2);(CO2);²- + H* Ka1 = 6.21 x 10; pKs1 = 4.207 K2 = 2.31 x 10; pK2 = 5.636

Appl Of Ms Excel In Analytical Chemistry

2nd Edition

ISBN:9781285686691

Author:Crouch

Publisher:Crouch

Chapter7: Neutralization Titrations And Graphical Representations

Section: Chapter Questions

Problem 10P

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Construct a curve for the titration of 20.00 mL 0.0500 N succinic acid (HO2CCH2CH2CO2H) by 0.1000 N NaOH solution. MS Excel® can be used. Titrant volumes: 15.00 mL, 20.00 mL, and 25.00 mL. (CH2)2(CO2H)2 ⇌ (CH2)2(CO2H)(CO2)^− + H^+ Ka1 = 6.21 × 10^-5; pKa1 = 4.207 (CH2)2(CO2H)(CO2)− ⇌ (CH2)2(CO2)2^2− + H^+ Ka2 = 2.31 × 10^-6; pKa2 = 5.636
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