Question

10. If you titrate 10.0 mL of 0.25 M morpholine (C_H,ONH, PK. = 5.51) with 0.10 M HNO3, what would be the pH after addition of 20.0 mL of HNO, solution? (a) 3.06 (b) 7.89 (c) 8.49 (d) 9.09 (e) 10.94 (f) none of the above

Answer 1

GHEONH (aq) + HNO (aq) + CAMEONH,O (.9+.NO, ( agy 10 mL of 0.25M CAHONH = 10 m/ * 0.25 molt' = 2.5 mmol of Cateo NH 20 mL of 0.10M ANO, = 20 mkx 0.10 mollt' 9 mmol of HNOL, Noo, I mol of HNO3 reacts with imol of CH, ONH to produce - 1 mol of Ca H₂ONHO. So, 2 mmol of thing will react with ammol of Cg H₂ONH to produce - 2 mool of CH₂ ONHO bo, amount of salt,cn Hg OM = 2 mool & amount of eneess hase , GH₂ ONE = 2.5mmolana mmol = 0.5 mmol Total volume = 10mL + 20 mL =30ml sa, coneen tration of ca ontemCCH, Ometico y = 24 mol = om t concentration of cq te ont, [exty onu) = 0,5 Miel LOM This is an renample of basic buffer. Hendes fon Hasselbalch equations polt = pk + dog, salt Il Given, bl=5.51 =5551+log PCA HG ON HET ICH ONET'

of box= 5.51+ login Pa 31 → po= 5.51 +logia (4) pot = 6:112059991 blz 14- poh=14–61112059991 = 7887940009-7189 07.89