Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Can someone help me with the calculations for Trail #1 for BOTH A and B? Thank y

ID: 1031517 • Letter: C

Question

Can someone help me with the calculations for Trail #1 for BOTH A and B? Thank you in advance

Sample no. Monoprotic or diprotic acid? monopronc Trial 1 1. Molar concentration of NaOH (mo/L) 2. Volume of weak acid (mL) 3. Buret readng of NaOH, initial (mL) 4. Buret reading NaOH at stoichiometric point, 0.09 25.005 0.00 O.C final (mL) 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of pH vs. VNaoH graph 7. Moles of NaOH to stoichiometric point (mol) 8. Moles of acid (mol) 9. Molar concentration of acid (mo/L) 10. Average molar concentration of acid (moVL) B. Molar Mass and the pK, of a Solid Weak Acid Sample no. 33 0134 on Monoprotic or diprotic acid? monopron Trial 1 1. Mass of dry, solid acid (8) 2. Molar concentration of NaOH (mol/L) 3. Buret readng of NaOH, initial (mL) 4. Buret reading NaOH at stoichiometric point final (mL) 5. Volume of NaOH dispensed (mL) 6. Instructor's approval of pH versus VsOH graph 7. Moles of NaOH to stoichiometric point (mol) 8. Moles of acid (mol) 9. Molar mass of acid (g/mol) 10. Average molar mass of acid (g/mol) 11. Volume of NaOH halfway to stoichiometric point (mL) 12. pKai of weak acid (from graph) 13. Average pKat Show calculations for Trial 1 on the next page. mass of weighing paper(g) mass or cighing papcr 0.9i5

Explanation / Answer

part A)trial 1)

Molarity of NaOH=0.099mol/L

Volume of weak acid=25.00ml

volume of NaOH dispensed=32.07

As monoprotic acid reacts with NaOH in 1:1 molar ratio,

mol of weak acid=mol of NaOH=M(NaOH)*V(NaOH)=0.099mol/L*32.07ml*(1L/1000ml)=0.00317 mol

molar concentration of acid=mol/vol=0.00317mol/0.025L=0.127mol/L

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote