Grades for Andrea Brisby: f. University of Oklahoma + O saplingleaming.com/ibisc
ID: 555541 • Letter: G
Question
Grades for Andrea Brisby: f. University of Oklahoma + O saplingleaming.com/ibiscrms/mod/ibisview.php?id=3871 621 Sapling Learning macmilan learning Jump to.. Andrea Brisby 1 2100 3 of 15 Available From: Due Date: Points Possible: Grade Category: 11/3/2017 06:00 PM 11/8/2017 11:55 PM 5 Homework University Science Books preeented tby Saplng Leaming lculate the pH of the resulting solution if 27.0 mL of 0.270 M HCI(aq) is added to 4 100 (a) 32.0 mL of 0.270 M NaOH(aq). attempt, 0% penalty. Policies: Solutions After Due Date Number 6 5100 pH12.63 You can check your answers. You can view solutions after the due date You have five attempts per question. There is no penalty for incorrect answers. (b) 37.0 mL of 0.320 M NaOH(aq). pH1.15 0 eTextbook 10 2 11 2 12 100 13 14 2 OHelp With This Topic O Web Help & Videos 100 O Technical Support and Bug Reports Previous #Try Again Next Exit 15 1 Explanation © 2011-2017 Sapling Learning, Inc. about u | careers | privacy policy | tams of use | contact us | help 11:59 PM Type here to searchExplanation / Answer
a)
we have:
Molarity of HCl = 0.27 M
Volume of HCl = 27 mL
Molarity of NaOH = 0.27 M
Volume of NaOH = 32 mL
mol of HCl = Molarity of HCl * Volume of HCl
mol of HCl = 0.27 M * 27 mL = 7.29 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.27 M * 32 mL = 8.64 mmol
We have:
mol(HCl) = 7.29 mmol
mol(NaOH) = 8.64 mmol
7.29 mmol of both will react
remaining mol of NaOH = 1.35 mmol
Total volume = 59.0 mL
[OH-]= mol of base remaining / volume
[OH-] = 1.35 mmol/59.0 mL
= 0.0229 M
we have below equation to be used:
pOH = -log [OH-]
= -log (2.288*10^-2)
= 1.64
we have below equation to be used:
PH = 14 - pOH
= 14 - 1.64
= 12.36
Answer:12.36
b)
we have:
Molarity of HCl = 0.27 M
Volume of HCl = 27 mL
Molarity of NaOH = 0.32 M
Volume of NaOH = 37 mL
mol of HCl = Molarity of HCl * Volume of HCl
mol of HCl = 0.27 M * 27 mL = 7.29 mmol
mol of NaOH = Molarity of NaOH * Volume of NaOH
mol of NaOH = 0.32 M * 37 mL = 11.84 mmol
We have:
mol(HCl) = 7.29 mmol
mol(NaOH) = 11.84 mmol
7.29 mmol of both will react
remaining mol of NaOH = 4.55 mmol
Total volume = 64.0 mL
[OH-]= mol of base remaining / volume
[OH-] = 4.55 mmol/64.0 mL
= 0.0711 M
we have below equation to be used:
pOH = -log [OH-]
= -log (7.109*10^-2)
= 1.15
we have below equation to be used:
PH = 14 - pOH
= 14 - 1.15
= 12.85
Answer: = 12.85
Feel free to comment below if you have any doubts or if this answer do not work
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.