O ACIDS AND BASES Calculating the pH of a weak base titrated with a strong acid
ID: 706293 • Letter: O
Question
O ACIDS AND BASES Calculating the pH of a weak base titrated with a strong acid An analytical chemist is titrating 199.3 mL of a 0.6300 M solution of diethylamine ((C2Hs)NH with a 0.7500 M solution of HNO3. The pK, of diethylamine is 2.89. Calculate the pH of the base solution after the chemist has added 192.5 mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added Round your answer to 2 decimal places. pH-Explanation / Answer
no of moles of (C2H5)2NH = molarity * volume in L
= 0.63*0.1993 = 0.125559moles
no of moles of HNO3 = molarity * volume in L
= 0.75*0.1925 = 0.144375moles
The no of moles of H^+ excess = 0.144375-0.125559 = 0.018816 moles
The total volume of solution = 199.3 + 192.5 = 391.8 ml = 0.3918 L
The concentration of H^+ ions = no of moles/ volume in L
= 0.018816/0.3918 = 0.048 M
[H^+] = 0.048 M
PH = -log[H^+]
= -log0.048 = 1.3187>>>>answer
please any doubt comment on the BOX
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