Question

An analytical chemist is titrating 241.3 mL of a 0.5300 M solution of diethylamine H with a 0.6500 M solution of HNO3. The pK, of diethylamine is 2.89. Calculate the pH of the base solution after the chemist has added 217.6 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 HNO2 solution added. Round your answer to 2 decimal places. pH = 1 xs ?

Answer 1

The question is solved using basic concepts of buffer formation.

WPAGE NO.: Solution - The given data is - volume of (ccants), NH) z 241-3 ml Concentration of (CGHS), NH) z 0:5300M Concentruotfon of HNOS. z 0:6500M pky far diethylamine z 2.89 milimales of (CG2 MB )NA) z_ 24:3 ml x 0.5300M - 127.89 milimales then millimales when 217.6 ml of the HNO3 added of HNOZ = 217 •6 X 0 · 6500 z +91-44 milimeler We can see that milimales of (CGHsh NM) are the limited hence - The milimales of salt form = 14.49 127.89 z 13.55 milimales at this stage - In solution there is only milimales of salt there is no milimale of (ccants), NH). Now pH of solution can be calculated as pH = 7 - 1 (Pxy + log (salt) Balt] 2 milimales of salt total volume PO10

DATE: 1 PAGE NO.: [salt) z 13:55 iilimal __(241-3 +217-6) mililiter [Seltj z 13.55 458.9 z 2.95X102 mole liten [salt]z 2.95 x10² Molar PH z 7 - E (2.89 +log (2.95 x162) ÞH 2_7-(2-89 +log 2.95 – 2 logto) b427-4 (2-89 + 0.4698 – 2) tz 7-4 (13598) bK27 - 0:6799_z 6-32 PH z 6.32