Youssef Abdeldayem (Student) Main Menu Contents Grades Reservations Course Conte
ID: 1081022 • Letter: Y
Question
Youssef Abdeldayem (Student) Main Menu Contents Grades Reservations Course Contents.... Review Assignment (Due 11:30pm Thursday, February 1, 2018) Determine the pH (to two decimal places) of the following solutions. A table of pka values can be t 5.81×10-3 M ethylammonium chloride 6.55 You are correct. Your receipt no. is 162-1173 3.78×10-3 M phosphoric acid[ Submit Answer Incorrect. Tries 1/3 Previous Tries 7.0Sx102 M methylamine Submit Answer Tries 0/3 2.19x103 M sodium hydrogen phosphate Submit Answer Tries 0/3 This discussion is closed.Explanation / Answer
3.78 * 10^-3 M of H3PO4
H3PO4 is a weak acid and its dissociation is as shown below,
H3PO4 <----> H2PO4- + H+ ; K1 = 7.1 * 10^-3
H2PO4- <-----> HPO4 2- + H+ ; K2 = 6.2 * 10^-8
HPO4 2- <-----> PO4 3- + H+ ; K3 = 4.2 * 10^-13
since the dissociation contants K2 and K3 are very small compare to K1 hece we consider only K1 in the calculating of pH of the solutin since [H+] ion coming from mainly from k1.
pH = pKa -logC
pH = -log(7.1*10^-3) - log(3.78 * 10^-3)
pH = 4.571
7.05 * 10^-2 M methylamine
CH3NH2 + H+ <-----> CH3NH3+ ; K = 5 * 10^-4
pOH = pKb - log C
pOH = -log(5 * 10^-4) -log(7.05 * 10^-2)
pOH = 4.453
pH = 14 - pOH
pH = 14 - 4.453
pH = 9.547
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