M Common HW #10 (LAST! x fi Dezto.mheducation.com/hm.tpx :: Apps P Pandora EMAIL
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M Common HW #10 (LAST! x fi Dezto.mheducation.com/hm.tpx :: Apps P Pandora EMAIL Th Moodle Top Hat whentowork N Netflix webadvisor BOM MCHEM Connect 9 CALC chem Li rent EJEMBE West Virginia Histor @Sign In: Discover McWilliams- CHEM112 FA15: CHEM 112 Sections vy Hancock CHEMISTRY 03&E1; Common Hw #10 (LAST!) instructions I help Question 9 (of 16) Save & Exit Submit 9 value 6.25 points 1 out of 3 attempts Be sure to answer all parts. Assistance Find the pH of the equivalence point and the volume (mL) of 0.150 M HCl needed to reach the equivalence point in the titration of 65.5 mL of 0.234M NH3 Check My Work Question Help Report a Problem Volume mL HCI pH = 11:51 AMExplanation / Answer
Given,
Molarity of HCl = 0.15 M
Molarity of NH3 = 0.234 M
Volume of NH3 solution = 65.5 mL
=> Millimoles of NH3 in solution = 0.234 x 65.5 = 15.327
The reacton is,
HCl + NH3 ----> NH4Cl
At equivalence point,
Moles of HCl = Moles of NH3
=> Millimoles of HCl added = 15.327
Millimoles = Molarity x Volume (mL)
=> 15.327 = 0.15 x V
=> Volume of HCl = 102.2 mL
Millimoles of NH4Cl produced = 15.327
Volume of Solution = 65.5 + 102.2 = 167.7 mL
=> [NH4Cl] = 15.327 / 167.7 = 0.0914 M
We know that,
Ka for NH4+ = 5.75 x 10^-10
NH4+ ---> NH3 + H+
0.0914-X.....X........X
Ka = [NH3] [H+] / [NH4+]
=> 5.75 x 10^-10 = X^2 / (0.0914 - X)
=> X = 7.25 x 10^-6 M = [H+]
pH = - log [H+] = - log (7.25 x 10^-6) = 5.14
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