G Google? Da itou dXw econ Home Main G Ch 99 Study G de Solu AAC Organic wR Engl
ID: 1034651 • Letter: G
Question
G Google? Da itou dXw econ Home Main G Ch 99 Study G de Solu AAC Organic wR English to French talian, . In: 06 Fracas interactf-Sapling Learning Login P w Sapling Learning Corey E 43.1/1003/27/2018 08 03 AM Periodic Table Due Date Points Possible 100 Grade Category Graded 3/28/2018 11:59 PM an important compound in industry and agriculture 1.30 6.70 Homework You can check your answers You can view solutions when you complete or give up on any question You can keep trying to answer each question until you get it right or give up You lose 5% ofthe points valable to each answer in your question for each incorrect 6 0 any KOH (b) after addition of 25.0 mL of KOH 8 0 (C) after addition of 50.0 mL of KOH 10 1 100 (d) after addišion of 75.0 ml of KOHD 12 2 13 0 14 0 15Explanation / Answer
H3PO3 millimoles = 50 x 1.8 = 90
a) Before any addition of KOH :
H3PO3 ---------------------> H+ + H2PO3-
1.8 0 0
1.8- x x x
Ka1 = [H+][H2PO3-] / [H3PO3]
0.05 = x^2 / 1.8 - x
x = 0.276
[H+] = 0.276 M
pH = -log [H+] = -log [0.276]
= 0.559
pH = 0.56
b) after addition of 25.0 mL KOH
it is first half equivaelce point
pH = pKa1 = 1.30
pH = 1.30
2 ) addition of 50.0 mL KOH
it is first equivalence point
pH = 1/2 (pKa1 + pKa2)
pH =1/2 (1.30 + 6.70)
pH = 4.0
3) 75.0 mL KOH
it is seond half equivalece point
pH = pKa2
pH = 6.70
4) 100.0 mL KOH
it is second equivalece point
HPO3^-2 millimoles = 100 x 1.8 = 180
HPO3^-2 molarity = 180 / (50 +100) = 1.2 M
HPO3^-2 + H2O ------------------> H2PO4- + OH-
1.2 -x x x
Kb2 = x^2 / 1.2-x
5.01 x 10^-8 = x^2 / 1.2-x
x = 2.65 x 10^-4
[OH-] = 2.45 x 10^-4 M
pOH = -log[OH-] = -log (2.45 x 10^-4 )
pOH = 3.61
pH + pOH = 14
pH = 10.39
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.